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THE   CALCULATIONS 


OF 


ANALYTICAL   CHEMISTRY 


ft&<^^ 


THE  CALCULATIONS 


OF 


ANALYTICAL  CHEMISTRY 


BY 

EDMUND   H.    MILLER,   Ph.D. 

PROFESSOR   OF  ANALYTICAL  CHEMISTRY 

IN 

COLUMBIA  UNIVERSITY 


THIRD  EDITION 

REVISED  AND  ENLARGED 


Keto  York 
THE   MACMILLAN   COMPANY 

LONDON:   MACMILLAN  &  CO.,  Ltd. 
1915 

All  rights  reserve,  d 


Entered  according  to  Act  of  Congress,  in  the  year  1905 

By  EDMUND   H.    MILLER 

In  the  Ofl&ce  of  the  Librarian  of  Congress,  at  Washington,  D.  C. 


Press  of 

iHE  New  Era  Printing  Company 
Lancaster,  Pa. 


PREFACE. 


This  text-book  is  intended  for  use  in  scientific  schools  and 
colleges,  in  connection  with  courses  in  analytical  chemistry,  and 
aims  to  give  a  logical  treatment  of  the  calculations  required  of 
an  analyst  As  analytical  chemistry  is  so  closely  allied  with 
the  other  branches  of  chemistry,  many  of  the  calculations  given 
will  be  found  of  general  application.  The  matter  has  been 
selected  so  as  not  to  encroach  on  the  domain  of  physical  chem- 
istry but  is  intended  to  lead  up  to  the  study  of  that  subject. 

The  object  has  been  to  furnish  a  text-book,  which  shall 
give  the  necessary  information  concerning  those  important 
chemical  calculations,  which  every  student  should  thoroughly 
understand  before  taking  up  advanced  work.  The  method  of 
presentation  and  arrangement  differs  materially  from  the  old 
books  on  chemical  calculations.  Formulae  have  been  avoided, 
except  in  the  last  chapters,  so  that  the  student  shall  consider 
each  problem  individually  and  solve  it  from  a  knowledge  of 
chemical  laws  instead  of  substituting  in  formulae  for  different 
cases.  If  the  student  has  in  this  way  obtained  ability  to  apply 
the  laws,  he  can  readily  construct  a  formula  or  table  for  con- 
stant use. 

More  than  two  hundred  examples  are  given,  with  answers ; 
all  of  which  can  be  solved  by  arithmetic  or  algebra.  They 
have  been  made  as  practical  as  possible  in  order  to  connect 
chemical  laws  with   practical  work.     Numerous  tables,  which 


303:4 


yi,  PREFACE. 

have  been  carefully  verified,  will  be  found  at  the  back  of  the 
book. 

The  assistance  of  Prof  Freedman  of  the  University  of  Ver- 
mont on  Chapter  X.  and  of  Dr.  Joiiet  in  correcting  proof  and 
verifying  examples  is  gratefully  acknowledged.     Any  corrections 

or  suggestions  will  be  much  appreciated. 

E.   H.  M. 
Havemeyer  Hall, 

January^  igoo. 


PREFACE  TO  THIRD  EDITION. 


As  a  new  edition  of  this  book  has  become  necessary,  the 
opportunity  is  taken  to  make  a  thorough  revision  and  enlarge- 
ment. Experience  with  this  text-book  during  the  past  four  years 
has  shown  it  to  be  advisable  to  subdivide  the  chapter  on  normal 
solutions  and  to  make  the  explanations  more  full.  The  inter- 
national atomic  weights  for  1904  have  been  introduced  through- 
out the  book  and  many  new  examples  added. 

The  author  acknowledges  a  number  of  corrections  and  help- 
ful suggestions  and  the  assistance  of  Mr.  F.  V.  D.  Cruser  in 
recalculating  the  examples  and  in  correcting  the  proof 

E.   H.   M. 
Havemeyer  Hall, 

September^  igo^. 


TABLE  OF  CONTENTS. 


CHAPTER  I. 

Calculation  of  Chemical  Equivalents  and  Atomic  Weights. 

PAGE. 

Determination  of  Equivalents i 

Two  Equivalents  of  Copper ...  4 

Electrolytic  Determination  of  Equivalents 4 

Definition  of  Atomic  Weight 5 

Relation  of  Chemical  Equivalents  to  Atomic  Weights 6 

Examples 9 

CHAPTER  II. 

Calculation  of  Formula  and  Percentage. 

Empirical  Formula  from  percentage 14 

Formulae  of  Minerals 15 

Oxygen  Ratio 18 

Percentage  Composition  from  Formula 19 

Percentage  to  the  Dry  Basis 19 

Examples 23 

CHAPTER  III. 

Calculations  of  Mixtures  Having  A  Common  Constituent. 

Chlorine  and  Bromine  in  Mixed  Silver  Salts 26 

Sodium  and  Potassium  in  Mixed  Sulphates 28 

Bicarbonate  and  Carbonate  in  Commercial  Bicarbonate 29 

Formation  of  Mixtures  of  Definite  Composition  31 

Examples  3S 

CHAPTER  IV. 

Calculations  from  Equations. 

Quantitative  Meaning  of  Equations  36 

Weight  of  a  Constituent  in  a  Given  Weight  of  a  Compound 37 

Weight  of  a  Constituent  not  in  the  Substance  Weighed 39 

Quantity  of  Reagent  Necessary  to  Complete  a  Reaction 40 

Choice  of  Most  Economical  Reagent 42 

Examples 43 


Viii  TABLE  OF  CONTENTS. 

CHAPTER  V. 

PAGE. 

Calculation  and  Use  of  Factors. 

Calculation  of  Factors 46 

Use  of  Factors  in  Gravimetric  Analysis 47 

Use  of  Factors  in  Volumetric  Analysis 49 

Assay  Ton  System  $0 

Examples 5^ 

CHAPTER  VI. 

Calculations  of  Volumetric  Analysis. 

Part  I.     General. 

Discussion  of  Normal  Solutions 55 

Adjustment  of  Solutions 62 

Subtraction  of  Excess  Necessary  to  Affect  the  Indicator 64 

Examples 66 

Part  II.     Reactions  of  Precipitation  and  Saturation. 

Precipitation  Calculations 69 

Calculations  of  Alkalimetry  and  Acidimetry yo 

Examples 75 

Part  III.     Reactions  of  Oxidation  and  Reduction. 

Permanganate  Calculations jS 

Iodine  Calculations 83 

Reduction  by  Standard  Solutions   85 

Examples  87 

CHAPTER  VII. 

Calculations  of  Density  of  Solids  and  Liquids. 

Density  of  Solids 91 

Reduction  to  Density  at  4"  C 96 

Density  of  Liquids 96 

Dilution  to  a  Certain  Specific  Gravity , 98 

Changes  in  Specific  Gravity  Due  to  Temperature  99 

Volume  Occupied  by  Precipitates 100 

Examples 103 

CHAPTER  VIII. 

Calculations  of  Gases. 

Corrections  for  Temperature  and  Pressure 106 

Density  of  Gases 108 

Relation  between  Density  and  Molecular  "Weight 108 

Density  by  Methods  of  Dumas,  Meyer  and  Hofmann no 

Density  b^  Effusion Ii3» 


TABLE  OF  CONTENTS.  ix 

FACE. 

Correction  of  Weighings 114 

Calculations  of  Gas  Analysis 117 

Examples 121 

CHAPTER  IX. 

Calctjlations  of  Calorific  Power. 

Thermo-Chemical  Reactions  127 

Heats  of  Formation 128 

Units  of  Calorific  Power  130 

Calorific  Power  of  Solids  from  the  results  of  Ultimate  Analysis 132 

Calorific  Power  of  Liquids 135 

Relations  of  Units  used  for  Gases 136 

Calorific  Power  of  Gases  from  the  Results  of  Analysis 137 

Calorific  Power  from  Weight  of  Lead  Reduced 138 

Examples  140 

CHAPTER  X. 

Electric  and  Electrolytic  Calculations  for  Direct  Currents. 

Electrical  Units  142 

Calculation  of  Resistance 143 

Ohm's  Law 145 

Joule's  Law 146 

Mechanical  and  Electrical  Units  of  Power 148 

Electrolysis 148 

Cathions  and  Anions 149 

Faraday's  Laws..  150 

Electro-chemical  Equivalents 151 

Formulae  for  Weight  of  Deposit,  Current  and  Power 151 

Counter-Electromotive  Force 153 

Thompson's  Law 155 

Ohm's  Law  for  Electrolysis 158 

Selective  Electrolysis 159 

Relations  between  Voltage  and  Current  in  Electrolytic  Work  161 

Effect  of  Dissolving  Anodes  162 

Current  Density 163 

Examples,  With  Solution  of  Each 164 

TABLES. 

Weights 175 

Measures 176 

Atomic  Weights '77 


X  TABLE   OF  CONTENTS. 

PAGE. 

Factors 178 

Specific  Heat  of  Solid  Elements 180 

Comparison  of  Centigrade  and  Fahrenheit  Degrees 181 

Values  of  Normal  Solutions 182 

Values  of  Tenth  Normal  Solutions 183 

Comparison  of  Degrees  Baume  with  Specific  Gravity 184 

Volvune  of  One  Gram  of  Water  at  Different  Temperatures 185 

Weights  of  One  Liter  of  Various  Gases 186 

Vapor  Pressure  of  Water 187 

Specific  Gravity  and  Percentage  of  Ethyl  Alcohol 188 

Percentage  and  Specific  Gravity  of  Sulphuric  Acid.. 189 

Percentage  and  Specific  Gravity  of  Nitric  Acid 190 

Specific  Gravity  and  Percentage  of  Hydrochloric  Acid 191 

Specific  Gravity  and  Percentage  of  Ammonia 192 

Relations  between  Units  of  Electricity,  Heat  and  Power 193 

Electro-Chemical  Equivalents 194 

Heats  of  Combination,  per  Gram-Equivalent 195 

Specific  Resistance  to  Electricity 196 

Calculated  and  Observed  Decomposition  Voltages 197 

Logarithms 198 

Antilogarithms , 200 


CHAPTER  I. 

CALCULATION     OF     CHEMICAL     EQUIVALENTS    AND    ATOMIC 

WEIGHTS. 

CHEMICAL  EQUIVALENTS. 

The  calculations  of  analytical  chemistry  are  based 
on  the  constant  relations  existing  between  the  different 
elements  in  a  chemical  compound.  These  relations  are 
best  expressed  by  equivalents,  that  is,  the  weight  of  the 
element  equivalent  to  one  part  of  hydrogen.  They  are 
the  equivalent  quantities  which  will  enter  into  reaction 
with  one  part  by  weight  of  hydrogen. 

For  example,  if  we  decompose  water  by  an  electric 
current,  we  have  hydrogen  and  oxygen  given  off  and  for 
every  gram  of  hydrogen  we  find  eight  grams  of  oxygen, 
therefore  the  equivalent  of  oxygen  is  eight,  for  eight 
parts  by  weight  of  oxygen  combine  with  one  part  by 
weight  of  hydrogen  to  give  nine  parts  of  water. 

The  equivalent  of  copper  may  be  obtained  as  follows: 
When  pure  black  oxide  of  copper  is  heated  in  a  stream 
of  hydrogen,  the  oxide  is  reduced  to  metallic  copper  and 
the  oxygen  combines  with  the  hydrogen  to  form  water, 
which  can  be  absorbed  by  a  weighed  calcium  chloride 
tube  and  reweighed  after  the  experiment. 

Suppose  we  take  four  grams  of  black  oxide  of  copper 
and  carry  on  the  experiment  until  no  more  water  is 
formed:  then  weigh;  the  metallic  copper  remaining  is 


2  CHEMICAL  EQUIVALENTS. 

3. 196  grams :  the  loss  in  weight  is  the  oxygen  which  was 
combined  with  the  copper ;  so,  as  we  know  the  equiva- 
lent of  oxygen  to  be  8,  if  we  make  the  proportion  3.196 
(the  weightofcopper):o.8o4(theweightof oxygen) : :  x:  8, 
X  will  be  the  equivalent  of  copper,  31.8.  Or  weigh  the 
water  formed  (the  increased  weight  of  the  calcium 
chloride  tube) :  this  is  0.9045  gram.  From  this  we  can 
obtain  a  direct  comparison  between  hydrogen  and  cop- 
per;  for,  of  the  0.9045  gram  of  water,  0.8040  gram  is 
the  oxygen  lost  by  the  copper  oxide,  so  0.9045-0.8040 
or  0.1005  gram  is  the  weight  of  hydrogen  which  has 
combined  with  the  oxygen  previously  in   combination 

with  copper;  hence  ^'^^    or  ^1.8  is  the  weiefht  of  copper 
0.1005  &  rr 

equivalent  to  one  gram  of  hydrogen  or  the  chemical 
equivalent  of  copper. 

To  obtain  the  equivalent  of  bismuth,  the  metal  can  be 
treated  with  nitric  acid  until  it  is  completely  converted 
to  oxide  and  after  ignition  weighed.*  3.6770  grams  of 
bismuth  gave  4.1016  grams  of  oxide.  The  increased 
weight  0.4246  gram  is  the  oxygen  combined  with 
3.6770  grams  of  bismuth,  hence  the  equivalent  of  bismuth 
will  be  found  from  the  proportion 

3  6770  :  0.4246 :  :  X :  8.  x  =  69.27. 

If  we  heat  finely  divided  bismuth  with  an  excess  of 
sulphur  and  add  small  portions  of  sulphur  until  the  bis- 
muth is  completely  converted  to  sulphide,  then  remove 
the  excess  of  sulphur  by  digesting  with  a  caustic  alkali 
solution,  we  can  from  the  increased  weight  find  the 
equivalent  of  sulphur.  For  example,  3  grams  of  bismuth 
gave  3.6928  grams  of  sulphide.  To  get  the  equivalent 
of  sulphur.     3  :  0.6928  ::  69.27  :  x.     x  =  16. 

*  Schneider,  J.  prakt,  Chem.  [2],  50,461. 


CHEMICAL  EQUIVALENTS.  3 

We  have  now  the  following  equivalents:  Oxygen,  8; 
Copper,  31.8;  Bismuth,  69.27;  Sulphur,  16.  To  obtain 
that  of  silver,  heat  together  pure  finely  divided  silver 
and  an  excess  of  sulphur,  volatilize  the  excess  of  sulphur 
and  weigh  the  resulting  silver  sulphide.  Suppose  these 
to  be  the  results  :  4  grams  of  silver  gave  4.592  grams  of 
sulphide ;  then  0.592  :  4 :  :  16  :  x.  x  =  108,  the  equivalent 
of  silver.  Now  that  we  have  obtained  the  equivalent 
of  silver,  we  can  combine  it  with  a  number  of  elements 
and  get  their  equivalents.  For  example,  dissolve  three 
grams  of  silver  in  nitric  acid  and  pass  in  chlorine  until 
the  silver  is  completely  precipitated  or  pass  chlorine  over 
the  silver  heated  in  a  combustion  tube  until  it  is  com- 
pletely converted  into  chloride:  weigh  the  silver  chloride; 
weight,  3.9861  grams.  To  calculate  the  equivalent  of 
chlorine  we  have  0.9861  :  3  :  :  x  :  108.  x  =  35.5.  In  a 
similar  way  the  equivalents  of  bromine  and  iodine  can 
be  found. 

This  process  of  obtaining  one  equivalent  from 
another  can  be  carried  on  indefinitely  until  those  of  all 
the  known  elements  have  been  determined,  but  each 
successive  comparison  involves  an  increase  of  error 
so  that  whenever  possible  these  indirect  results  are  con-^ 
firmed  by  a  direct  comparison  with  hydrogen  or  oxygen. 
In  the  course  of  such  a  set  of  determinations,  it  fre- 
quently occurs  that  two  or  more  entirely  different 
equivalents  are  obtained  for  the  same  element. 

This  shows  that  the  element  may  combine  in  differ- 
ent proportions  and  as  the  different  equivalents  are  in- 
variably multiples  of  the  lowest,  it  proves  that  the 
elements  combine  in  simple  and  definite  proportions 
represented  by  whole  numbers. 

To  illustrate  this :  suppose  in  the  experiment  des- 
cribed  with  black  oxide  of  copper,    we  substitute    red 


4  CHEMICAL  EQUIVALENTS. 

oxide — 4  grams  as  before,  and  weigh  the  copper  remain- 
i"gf — 3-553  grams.  We  shall  have  to  obtain  the  equiv- 
alent of  copper 

3.553  :  :  0.447  : :  X  :  8.  x  =  63.6, 
which  is  exactly  double  that  obtained  from  the  black 
oxide,  so  that  8  parts  by  weight  of  oxygen  combine  in 
one  case  with  31.8  parts  of  copper  and  in  the  other  with 
63.6  parts.  This  leads  to  the  conclusion  that  the  red 
oxide  contains  exactly  one  half  as  much  oxygen  as  the 
black  oxide  in  proportion  to  copper.  When  we  attempt 
to  construct  formulae  from  these  results,  we  are  con- 
fronted by  a  difficulty,  for  CuoO  and  CuO  or  CuO  and 
CuOo  equally  well  represent  the  proportions  of  oxygen 
to  copper  found.  To  distinguish  between  these,  both 
molecular  weight  determinations  and  study  of  the  salts 
corresponding  to  the  respective  oxides  and  the  analogous 
compounds  of  similar  elements,  are  necessary. 

Equivalents  may  also  be  determined  electrolytically, 
for  according  to  Faraday's  Law,  the  same  quantity  of 
electricity  precipitates  or  liberates  chemically  equivalent 
quantities  of  all  elements.  Suppose  we  connect  in  series, 
cells  containing  suitable  solutions  of  cuprous  copper, 
cupric  copper  and  of  silver  respectively,  each  provided 
with  platinum  electrodes;  then  pass  a  weak  current 
through  the  three,  as  they  are  connected  in  series  they 
obtain  the  same  current;  so  that  the  weights  of  metal 
precipitated  are  proportional  to  the  chemical  equivalents. 
Suppose  the  weights  precipitated  are  silver  0.756  gram ; 
copper,  0.4452  in  one  case,  and  0.2226  in  the  other.  If 
we  know  one  equivalent  we  can  find  the  others ;  if  we 
Know  that  of  silver  to  be  108,  to  find  the  two  equivalents 
of  copper  we  have 

0.756  :  0.4452  :  :  108  :  X.     x  =  63.6. 

0.756  :  0.2226:  :  108  :  y.     y  =  3i.8. 


ATOMIC   WEIGHTS.  I 

For    the    relations   between    chemical  and  electro- 
chemical equivalents  see  Chapter  X. 


ATOMIC    WEIGHTS. 


The  atomic  weight  of  an  element  is  the  weight  of 
an  atom  compared  with  the  weight  of  an  atom  of  hydro- 
gen taken  as  unity  or  an  atom  of  oxygen  taken  as  six- 
teen. This  latter  definition  is  very  generally  adopted, 
for  the  ratio  of  oxygen  to  hydrogen  has  been  found  by 
most  careful  experiments  to  be  not  i6  to  i  but  15.88  to 
I  and,  as  the  value  for  oxygen  enters  into  a  great  num- 
ber of  atomic  weight  calculations,  it  is  more  convenient 
to  take  it  as  a  whole  number,  16,  and  make  it  the  base 
of  the  system.  If  we  use  O  =  16,  hydrogen  then  becomes 
1.008  when  great  exactness  is  required. 

The  atomic  weight  is  equal  to,  or  a  multiple  of,  the 
equivalent,  for  an  element  can  combine  with  one  or  more 
hydrogen  atoms  or  some  other  element  in  different  pro- 
portions depending  on  its  valence,  and  while  this  affects 
its  equivalent  it  does  not  its  atomic  weight  which  is  an 
invariable  property  of  the  element.  To  determine  which 
multiple  of  the  equivalent  is  the  atomic  weight  is  some- 
times a  matter  of  difficulty ;  which  requires  a  knowledge 
of  the  laws  and  methods  of  physical  chemistry  for  its 
certain  solution,  particularly  those  regarding  molecular 
weights,  which  are  not  within  the  province  of  this  book. 
If,  however,  we  take  the  least  common  multiple  of  the 
equivalents,  when  the  element  has  several,  we  get  a 
number  which  satisfies  all  analytical  data  and  is  the 
atomic  weight  subject  to  confirmation  by  physical  data 
and  the  periodic  law. 

In  order  not  to  leave  the  question  of  atomic  weights 
in  such  a  vague  condition,  let  us  assume  the  correctness 


6  ATOMIC  WEIGHTS. 

of  the  following  facts,  for  the  proof  of  which  the  student 
is  referred  to  books  on  physical  chemistry. 

1.  That  the  product  of  the  specific  heat  of  an  ele- 
ment in  the  solid  state  and  the  atomic  weight  is  a  con- 
stant (approximately)  6.4.     Law  of  Dulong  and  Petit. 

2.  That  equal  volumes  of  all  gases  under  the  same 
temperature  and  pressure  contain  the  same  number  of 
molecules.     Law  of  Avogadro. 

3.  That  the  gaseous  molecules  of  all  elements  con- 
sist of  two  atoms,  except  potassium,  sodium,  zinc,  cad- 
mium, and  mercury  which  are  monatomic,  phosphorus 
and  arsenic  which  are  tetratomic  and  sulphur  which  is 
variable  according  to  temperature.  (Phosphorus,  arsenic 
and  sulphur  are  diatomic  however  at  very  high  temper- 
atures). 

With  the  aid  of  these  laws  we  can  readily  confirm 
our  provisional  atomic  weights  as  follows: — The  least 
common  multiple  of  our  equivalents  for  copper,  31.8  and 
63.6,  is  63.6;  the  specific  heat  of  copper  is  0.095  (see 
table)  63.6  X  0.095  =6.042  ;  a  result  which  approximates 
6.4  with  sufficient  closeness  to  show  any  other  multiple 
of  the  equivalent  to  be  incorrect.  Again  we  found  the 
equivalent  of  silver  to  be  108;  its  specific  heat  is  0.057 
and  the  product  is  6. 1 5,  a  result  which  shows  that  the 
atomic  weight  and  equivalent  of  silver  are  identical. 

This  principle  can  be  applied  to  all  elements  whose 
specific  heats  in  the  solid  state  are  accurately  known, 
but  will  not  aid  us  to  confirm  the  atomic  weight  of  many 
of  the  non-metallic  elements.  To  do  this  we  must  con- 
sider the  other  facts.  Take  a  diatomic  molecule  such 
as  chlorine  and  compare  the  weight  of  a  liter  of  this 
with  the  weight  of  a  liter  of  hydrogen,  at  the  same  tem- 
perature and  pressure.  According  to  the  law  of 
Avogadro  these  volumes  contain  the  same  number  of 


ATOMIC  WEIGHTS.  7 

molecules.  Let  x=number  of  molecules  in  a  liter,  then 
weight  X  chlorine  molecules :  weight  x  hydrogen  mole- 
cules: :  weight  one  chlorine  molecule :  weight  one  hydro- 
gen molecule.  And  as  one  chlorine  molecule  and  one 
hydrogen  molecule  each  contain  two  atoms,  we  have, 
weight  liter  of  chlorine :  weight  liter  of  hydrogen : : 
atomic  weight  of  chlorine :  atomic  weight  of  hydrogen : 
or  3.1673  :  0.0896: :  X  :  I.  x=35.35.  A  result  which  con- 
firms with  sufficient  accuracy  our  equivalent  35.5  as  the 
atomic  weight  of  chlorine, 

To  determine  the  atomic  weight  of  mercury, 

1.  Ignite  5  grams  of  mercuric  oxide  and  weigh  the 
resulting  mercury — 4.62  grams.  The  loss  in  weight 
0.38  gram  is  oxygen  and  the  equivalent  of  oxygen  is  8 
so  0.38  :  4.62  :  :  8  :  x.  x  =  97.26.  The  equivalent  of  mer- 
cury. 

2.  Dissolve  5  grams  of  corrosive  sublimate  in  water, 
acidify  with  nitric  acid  and  precipitate  the  chlorine  in 
combination  by  silver  nitrate.  Weigh  the  silver  chloride, 
5.2952  grams.  We  have  found  the  equivalents  of  chlor- 
ine and  silver  to  be  35.5  and  108,  therefore  to  get  the 
chlorine  in  this  precipitate  we  have 

108+35.5  :  35.5  :  :  5.2952  :  x.      x  =  1.3099  grams, 
the  amount  of  chlorine  combined   originally  with   the 
mercury;   5 — 1.3099  =  3.6901,   the  amount  of  mercury 
present  in   the  five  grams  of  mercuric  chloride.     So  to 
get  the  equivalent  of  mercury  we  have, 

3.6901  :  1.3099  •  •  X  :  35-5-     x  =  100.+ 

3.  Fuse  5  grams  of  calomel  with  an  excess  of  alka- 
line carbonate  in  a  porcelain  crucible,  leach  out  the 
alkaline  chloride  formed  and  precipitate  it  with  an  excess 
of  silver  nitrate  ;  weigh  the  silver  chloride,  3.0466  grams. 


8  ATOMIC  WEIGHTS. 

Calculate  the  chlorine  present  as  before 

143-5  •  35-5  •  •  3-0466 :  X.     x  =  0.7537  gi'am. 

To  calculate  the  equivalent  of  mercury  we  have 

5-  —  0.7537:  0.7537  ::x:  35-5-     x  =  200. 

We  have  before  us  three  equivalents  of  mercury,  between 
the  first  and  second,  the  indirect  determination  has  the 
greater  probability  of  accuracy  on  account  of  loss  of 
mercury  by  heating,  so  the  two  reliable  equivalents 
are  100  and  200.  Therefore  the  provisional  atomic 
weight,  the  least  common  multiple,  is  200.  This  is  easily 
confirmed  by  the  specific  heat:  0.0319  times  200  =  6.38. 
So  that  200  is  the  atomic  weight  of  mercury  and  we  can 
assign  to  the  two  chlorides  formulae  of  HgCl  and  HgCla 
unless  their  molecular  weights  show  them  to  be  multi- 
ples. 


EXAMPLES. 


For  specific  heat  of  metals  see  tables  on  page  1 80. 

1.  Five  grams  of  pure  zinc  were  treated  with  dilute 
hydrochloric  acid  and  the  hydrogen  evolved  was  dried, 
measured  and  the  corresponding  weight  calculated;  this 
was  0.1529  gram.  What  is  the  equivalent  and  the 
atomic  weight  of  zinc?  Ans.     32.7  and  65.4. 

2.  One  gram  of  pure  metallic  magnesium  was  burned 
in  an  atmosphere  of  oxygen  and  the  resulting  oxide 
collected  and  weighed.  Weight  1.658  grams.  What 
is  the  atomic  weight  of  magnesium  ?        Ans.     24.31. 

3.  a.  Two  grams  of  silver  were  dissolved  in  nitric 
acid  and  bromine  water  added  until  no  more  precipitate 
formed.  Weight  of  silver  bromide  was  3.481  grams. 
Given  the  equivalent  of  silver,  108.  What  is  the  equiva- 
lent of  bromine  ?  Ans,     79.97. 

b.  Given  the  weight  of  a  liter  of  bromine  vapor, 
7. 1675  grams  and  the  weight  of  a  liter  of  hydrogen  under 
the  same  conditions,  0.0896  gram.  What  is  the  atomic 
weight  of  bromine  ?  Ans.     79.99. 

4.  Three  grams  of  silver  are  precipitated  by  an  ex- 
cess of  iodine.  Weight  of  silver  iodide  was  6.5277 
grams. 

Three  grams  of  silver  are  precipitated  by  an  excess 
of  chlorine.     Weight  of  silver  chloride,  3.9861  grams. 

Given  the  equivalent  of  chlorine,  35.5.  What  is  the 
equivalent  of  iodine  ?  Ans.     127. 

(9) 


10  EXAMPLES. 

5.  Roscoe  found  that  6.4428  grams  of  Cape  dia- 
monds gave  23.6275  grams  of  carbon  dioxide  when 
burned  in  air. 

Van  der  Plaats  found  that  30.8891  grams  of  sugar 
charcoal  gave  113.2320  grams  of  carbon  dioxide.  Given 
0  =  16.     What  is  the  atomic  weight  of  carbon  ? 

Ans.     12. 

6.  Richards  and  Rogers  found  that  2.35079  grams 
of  zinc  bromide  gave  3.91 941  grams  of  silver  bromide. 
Given  Br  =  79.95  and  Ag  =  107.92.  What  are  the 
equivalent  and  atomic  weights  of  zinc  ? 

Ans.     32.73  and  65.46. 

7.  Winkler  converted  electrolytically  deposited 
nickel  to  a  dry  chloride  and  then  weighed  the  chlorine 
as  silver  chloride.  He  obtained  from  0.301 1  gram  of 
nickel  1.462 1  grams  of  silver  chloride.  Given  Ag  = 
107.92  and  CI  =  35.37.  What  are  the  equivalent  and 
atomic  weights  of  nickel  ?  Ans.     29.5  and  59. 

8.  Deville  found  that  0.6763  gram  of  boron  chlor- 
ide, obtained  by  the  action  of  gaseous  hydrochloric  acid 
on  boron,  gave  2.4770  grams  of  silver  chloride.  Given 
Ag  =  107.92  and  CI  =  35.37.  What  is  the  chemical 
equivalent  of  boron  ?  Ans.     3.75. 

9.  Dumas  found  that  it  required  4.168  grams  of 
silver  (as  nitrate)  to  precipitate  the  chlorine  from  4.0162 
grams  of  anhydrous  barium  chloride.  Given  the  equiva- 
lents of  silver  and  chlorine,  107.92  and  35.37.  What  is 
the  chemical  equivalent  of  barium  ?  Ans.     68.62. 

10.  Stas  found  that  after  adding  7.25682  grams  of 
potassium  chloride  to  10.51995  grams  of  silver,  dissolved 
in  nitric  acid,  that  0.0194  gram  of  silver  remained  in 
solution.  Given  the  atomic  weights  of  chlorine  and 
silver  as  35.37  and  107.92.    Calculate  that  of  potassium. 

Ans.     39.2. 


EXAMPLES.  11 

11.  Mallet  determined  the  atomic  weight  of  alumi- 
nium by  the  following  methods  : 

(a)  by  oxidizing  to  water  the  hydrogen  evolved  when 
aluminium  was  dissolved  by  sodium  hydroxide  :  5.2632 
grams  of  aluminium  gave  5.2562  grams  of  water. 

(b)  by  precipitating  the  bromine  in  aluminium  bro- 
mide as  silver  bromide  :  8.6492  grams  of  aluminium  bro- 
mide required  10.4897  grams  of  silver. 

Calculate  the  atomic  weight  of  aluminium.  Given 
Ag  =  107.92,     Br  =  79.95,    O  =  16,  H  =  1.008. 

Ans.     (a)  2  7.06.     (b)  2  7. 105. 

12.  If  5  grams  of  a  metallic  bromide  gave  7.2552 
grams  of  iodide,  and  the  atomic  weights  of  bromine  and 
iodine  are  79.95  and  126.85  respectively.  What  would 
be  the  atomic  weight  of  the  metal  if  it  were  univalent, 
bivalent  or  trivalent  ?  Ans.     24,48,72. 

13.  1.4480  grams  of  calcite  gave  on  ignition  0.8 11 6 
gram  of  lime.  Given  the  atomic  weights  of  carbon  and 
of  oxygen  as  1 2  and  1 6.  What  is  the  chemical  equiva- 
lent of  calcium  ?  A71S.     20.057. 

14.  2.6146  grams  of  metallic  tin  gave  3.3196  grams 
of  oxide.  Given  the  chemical  equivalent  of  oxygen  as  8. 
What  is  the  atomic  weight  of  tin  ?  Ans.     1 18.68. 

.  15.  0.5302  gram  of  metallic  antimony  was  mixed 
with  sulphur  in  a  porcelain  boat  and  heated  at  250°  C.  in 
a  current  of  carbon  dioxide  until  the  excess  of  sulphur 
was  sublimed.  The  resulting  sulphide  weighed  0.7429 
gram.  Given  the  equivalent  of  sulphur,  16.  What  is  the 
atomic  weight  of  antimony  ?  Ans.     1 19.64. 

16.  o.  1 700  gram  of  zinc  gave  on  treatment  with  acid 
58.23  c.c.  of  hydrogen  gas.  Given  the  weight  of  a  litre 
of  hydrogen,  0.0896  gram.  What  is  the  atomic  weight 
of  zinc?  ^^^s.     65.16. 


12  EXAMPLES. 

17.  If  0.0899  gram  of  aluminium  gave  110.8  c.c.  of 
hydrogen,  what  is  the  atomic  weight  of  aluminium  ? 
Given  the  weight  of  a  litre  of  hydrogen,  0.0896  gram. 

Ans,     27.15. 

18.  0.5526  gram  of  metallic  bismuth  showed  an 
increase  in  weight  of  0.1278  gram  on  being  completely 
converted  to  sulphide.  Given  the  equivalent  of  sulphur 
1 6.     What  is  the  atomic  weight  of  bismuth  ? 

Ans.     207.54. 

19.  If  4  grams  of  an  oxide  gave  5  grams  of  a  sulphide, 
and  the  atomic  weights  of  oxygen  and  sulphur  are  1 6  and 
32,  what  would  be  the  atomic  weight  of  the  element, 
if  it  were  univalent,  bivalent  or  trivalent  ? 

Ans.     24,  48  and  72. 

20.  Parsons  found  that  1.4793 1  grams  of  basic 
beryllium  acetate,  Be40  (C2H302)6>  gave  0.36534  gram 
of  oxide,  BeO.  Calculate  the  atomic  weight  of  Be. 
Given  C  =  12,  O  =  16,  H  =  1.008.  Ans.     9.101. 

21.  Scott  found  that  4.84099  grams  of  ammonium 
bromide  (NH4Br),  when  precipitated  by  silver  nitrate, 
gave  5.33177  grams  of  silver  Given  H  =  1.008,  Br 
=  79.96,  Ag  =  107.93.  What  is  the  atomic  weight  of 
nitrogen?  Ans.     14.00.   , 

22.  1.05462  grams  of  silver  arsenate  (Ag3As04) 
were  converted  into  silver  chloride  by  passing  a  current 
of  hydrogen  chloride  gas  over  the  substance  in  a  porce- 
lain boat,  0.98014  gram  of  silver  chloride  was  obtained.* 
Given  Ag  =  107.93,  O  =  16,  CI  =  35.45.  What  is 
the  atomic  weight  of  arsenic  ?  Ans.     75.04. 

23.  Richards  and  Merigold  found  that  from  1.7999 
grams  of  uranium  bromide  (UrBr4),  when  converted  to 
AgBr,  1.39 1 8  grams  of  silver  were  obtained.     What  is 

*Ebaugh,  J,  Am.  C.  S.,  24,  493. 


EXAMPLES.  13 

the  atomic  weight  of  uranium?      Given  Ag  =  107.93; 
Br  =  79.96.  Ans.     238.47. 

24.  Lenher  found  that  0.98396  gram  of  pure  silver 
selenite  (AggSeOg),  yielded  0.82232  gram  of  silver 
chloride,  when  treated  in  a  current  of  hydrochloric  acid 
gas  to  expel  the  selenium  and  oxygen.  Given  Ag  = 
107.93,  0  =  16,  CI  =  35.45.  What  is  the  atomic 
weight  of  selenium  ?  Ans.     79.27. 

25.  Taylor  heated  0.48514  gram  of  metallic  tungsten 
in  a  current  of  oxygen  until  all  the  metal  was  converted 
to  oxide,  which  weighed  0.61 166  gram.  Given  the  chem- 
ical equivalent  of  oxygen  as  8.  What  is  the  atomic 
weight  of  tungsten  ?  Ans.     184.06. 

After  this  introduction  on  the  determination  of  atomic 
weights,  the  student  should  have  a  definite  idea  of  their 
meaning  and  how  they  can  be  obtained.  In  the  calcula- 
tions and  examples  from  this  point  the  atomic  weights 
given  on  page  177  will  be  used  unless  especially  men- 
tioned. These  were  adopted  by  the  International  Com- 
mittee on  atomic  weights  for  1904. 


CHAPTER    II. 

CALCULATION  OF  FORMULAE  AND  PERCENTAGE. 
FORMULA    FROM    PERCENTAGE. 

In  any  chemical  compound,  if  we  divide  the  percent- 
ages obtained  from  analysis  by  the  atomic  weights  of 
the  elements,  we  shall  get  numbers  which  represent  the 
ratio  of  atoms,  and  on  reducing  these  to  the  nearest 
whole  numbers,  obtain  the  empirical  formula — the  sim- 
plest ratio  of  atoms.  The  correct  formula  may  however 
be  a  multiple  of  this  which  would  have  the  same  percent- 
age composition.  Therefore  a  determination  of  the 
molecular  weight  is  necessary  to  decide  what  multiple  of 
the  empirical  formula  is  the  correct  one.  For  the  meth- 
ods and  calculations  used  in  determining  molecular 
weights,  except  from  the  specific  gravity  of  a  gas,  the 
reader  is  referred  to  the  numerous  text  books  on  phys- 
ical chemistry. 

The  empirical  formula  is  usually  obtained  from  an 
analysis,  and  while  the  proper  multiple  should  be  used 
in  all  equations,  yet  the  empirical  formula  serves  many 
useful  purposes  in  inorganic  analysis,  such  as  furnishing 
data  for  normal  solutions,  factors,  etc.,  just  as  correctly 
as  the  proper  molecular  formula. 

In  organic  chemistry,  this  is  not  the  case  as  the  ex- 
istence of  many  totally  different  bodies  having  the  same 
empirical  formula,  makes  it  essential  to  use  the  molecu- 

(14) 


FORMULAE  OF  MINERALS.  16 

lar  or  better  the  graphic  formulae.     The  following  ex- 
amples illustrate  the  calculation  of  empirical  formulae : 

A  salt  gave  on  analysis  52.45%  of  potassium  and 
47-55%  of  chlorine, 

52.45  -^39-15  =  1.340, 
47-55  ^  35-45  =  1-341, 
the  ratio  of  atoms  is  1.341  to  1.341  or  i  to  i,  so  the  for- 
mula has  an  equal  number  of  atoms  of  potassium  and 
chlorine  and  the  empirical  formula  is  KCl. 

A  salt  gave  on  analysis  the  following : 
Phosphorus         -  2 1.82%-- 31.       =0.704!    j^^^.^ 

Hydrogen  -  0.71%-f-i.oi     =0.703!     ^^ 

Sodium  -  32.43%  -^  23.05  =  1.407  \js^^^^^ 

Oxygen  by  difference  45.02% -^  16.       =  2. 814  J 

If  we  divide  the  numbers  representing  the  ratio  of 
atoms  by  the  greatest  common-divisor  0.703  we  get  the 
ratio  expressed  in  the  nearest  whole  numbers,  or  phos- 
phorus 1 ;  hydrogen  i ;  sodium  2  ;  oxygen  4.  The  em- 
pirical formula  is  therefore  PHNao04  or  as  it  is  usually 
written  NaoHP04. 

Analysis  of  ignited  magnesium  ammonium  phosphate 
gave  the  following  results  : 

Magnesium  21.82%-^  24.36  =  0.8957  -^  0.449  =  2. 
Phosphorus  27.86% -f-  31.  =  0.8987  -^  0.449  =  2. 
Oxygen        50.32% -f-  16.      =3-^45    -^0.449  =  7. 

The  greatest  common  divisor  of  the  numbers  repre- 
senting the  ratio  of  atoms  is  0.449  •  dividing  by  this  we 
obtain  the  nearest  whole  numbers.  So  the  empirical 
formula  is  Mg2P207. 

FORMULAE    OF    MINERALS. 

From  the  results  of  analysis  the  formula  of  a  mineral 
can  be  calculated  as  in  the  case  of  any  other  empirical 
formula ;   but  as  one  element  or  group  often  replaces 


16  FORMULAE  OF  MINERALS. 

another  to  a  greater  or  less  extent  without  altering  the 
mineralogical  character,  those  elements  which  may  re- 
place each  other  are  grouped  together  and  represented 
by  the  letter  R.  So  oxides  of  the  monatomic  elements, 
such  as  the  alkalies,  are  represented  by  RgO ;  oxides 
of  the  diatomic  elements,  such  as  calcium,  magnesium, 
barium,  ferrous  iron,  manganese,  etc.,  by  RO;  oxides  of 
aluminium,  chromium,  ferric  iron,  manganic  manganese, 
etc.,  by  R2O3  and  so  on.  In  this  way  the  composition 
of  the  mineral  is  indicated  and  especially  the  ratio  of 
basic  to  acid  oxides. 

For  example,  analysis  of  Franklinite  gave  the  follow- 
ing results  :* 


(0 

(2) 

(3) 

FegOs 

60.52 

66.34 

67.42 

MnsOa 

6.79 

ZnO 

19.44 

20.26 

6.78 

MnO 

12.81 

12.31 

9-53 

FeO 

15-65 

We  see  in  (i)  a  replacement  of  part  of  the  FeoOg  by 
MngOg;  so  that  R2O3  in  this  case  is  ferric  iron  and  man- 
ganic manganese,  comparing  (i),  (2),  and  (3),  we  see 
that  the  percentages  of  ZnO,  MnO  and  FeO  vary  greatly, 
these  constitute  the  RO  group. 

In  (i)  calculate  from  the  percentage  of  Mn203  the 
equivalent  percentage  of  FegOg.  To  do  this  we  have, 
the  percentage  of  Mn203  is  to  the  percentage  of  FeoOg 
as  the  molecular  weight  of  Mn203  is  to  the  molecular 
weight  of  Fe203.  The  molecular  weight  of  MngOg  is 
(2  X  55)  + 1  (3  X  16)  or  158  and  that  of  FesOe  is  (2 
X  55-9)  +  (3  X  16)  or  159.8.  So  the  proportion  becomes, 
6.79  :x::  158  :  159.8.     x  =  6.87%. 

*  Dana's  Mineralogy. 

I  In  these  calculations  the  atomic  weights  to  one  decimal  place  are  sufficient. 


FORMULAE  OF  MINERALS.  17 

In  the  same  way,  using  the  proper  molecular  weights, 
calculate  the  percentage  of  MnO,  to  the  equivalent  per, 
centage  of  ZnO, 

12. 8i  :  y  :  :  71.  :  81.4.     y  =  14.68%. 

We  have  now  FeoOg,  60.52%+  6.88%  =  67.40%  and 
ZnO,  19.44%  +  14-68%  =  34-12%.  Dividing  these  by 
the  molecular  weights  we  get 

67.40%^  159.8  =  0.422  -^0.420  =  I.  + 
34.12%-i-    81.4  =  0.419  -^  0.420  =  I.  — 

These  results  are  almost  identical,  so  the  ratio  of  RoOg 
to  RO  is  I  to  I  and  the  formula  is  represented  by 
RO,R203. 

There  was  no  particular  reason  for  calculating  the 
percentage  of  MnO  to  ZnO  rather  than  ZnO  to  MnO ; 
the  result  when  divided  by  the  proper  molecular  weight 
will  be  the  same  in  either  case. 

The  typical  formula  can  also  be  found  by  dividing 
the  percentage  of  each  oxide  by  its  molecular  weight 
and  then  grouping  together  the  ratios  as  follows — 

(2)     FeaOa  66.34  --  159.8  =  0.415. 

ZnO     20.26--    81.4  =  0.249]       ^ 
MnO    12.31 ->    71.    =0.172/        '^ 

The  ratio  is  approximately  one  to  one  as  before.  The 
lack  of  agreement  in  the  figures  representing  the  ratio 
is  probably  due  to  a  deficiency  in  the  percentage  of  R2O3 
elements  for  the  percentages  only  add  up  to  98.91. 

From  the  third  analysis  we  get  0.422  :  0.435,  again 
showing  the  ratio  to  be  one  to  one. 

So  from  the  results  of  these  three  analyses  we  find  the 
formula  of  Frankllnlte  to  be  (Zn  Mn  Fe)0,  (Fe  Mn)o03. 


18  FORMULAE  OF  MINERALS. 

Dana's  Mineralogy  gives  the  following  analyses  of 
garnets, 


(0 

(2) 

SiOj 

40.90 

39.85 

AL03 

22.81 

22.07 

Fe^Os 

I-I3 

CrgOa 

1.48 

FeO 

13-34 

MnO 

0.38 

MgO 

16.43 

0.68 

CaO 

4.70 

36.31 

The  R2O3  bases  are  calculated  to  AI2O3  as  the  per- 
centages of  Fe203  and  Cr203  are  small  and  indicate  a  par- 
tial replacement  of  the  AI2O3.  With  the  RO  bases  we 
may  calculate  either  to  CaO  or  MgO  and  divide  the  sum 
of  the  resulting  percentages  by  the  molecular  weight  of 
the  oxide  we  select.  After  calculating  Cr203  and  FesOg 
to  AI0O3  and  FeO,  MnO,  and  MgO  to  CaO  we  get 

('^ 

•^1^2  40.90%  -^    60.4  =  0.677.     3- 

Al203(R203)  23.80%  -^   102.2    =  0.236.         I. 

CaO  (RO)  38.25% -f-    56.1=0.682.     3. 

(2) 

Si02  39-85%-^    60.4  =  0.660.  3. 

Al2  03(R203)  i2.79%-^  102.2  =  0.223.  I. 

CaO(RO)  37.26%--    56.1=0.664.  3. 

So  in  both  cases  the  type  formula  is  3RO,  R2O3,  3Si02. 
In  connection  with  silicates  both  natural  and  artificial 
(slags)  the  term  "oxygen  ratio"  is  often  used.  This 
means  the  ratio  of  the  oxygen  atoms  belonging  to  the 
basic  groups  to  those  combined  with  silicon.  In  the 
garnet  we  have  six  oxygen  atoms  combined  as  basic 
oxides  and  six  as  an  acid  oxide  (SiOs),  hence  the  oxygen 


PERCENTAGE  COMPOSITION  FROM  FORMULA.  19 

ratio  is  one  to  one,  or  if  it  is  desired  to  distinguish  be- 
tween the  basic  oxides  1:1:2.  In  writing  these  ratios 
the  order  adopted  is  first  RO,  then  R2O3  and  last  SiOg. 


PERCENTAGE    COMPOSITION    FROM    FORMULA. 

If  we  have  the  formula  of  a  compound  whether  mole- 
cular or  empirical  and  the  atomic  weights  of  its  elements 
the  calculation  of  the  percentage  composition  is  made  as 
follows. — Take  for  example  sodium  chloride ;  we  have 
the  formula  NaCl  and  the  atomic  weights  Na  23.05  and 
CI  35.45.  The  molecular  weight  is  the  sum  of  the 
atomic  weights,  23.05  +  35-45  =  5B.5,  the  molecular 
weight  of  NaCl.     Let  x  =  %  Na  and  y  =  %  Q\  then 

23-05  :  58.5  *•  •  X  :  100.    X  =  39402%. 

35.45  :  58.5  :  :  y  :  100.    y  =  60.598%. 

To  prove  x  +  y  =  100.00%. 
To  calculate  the  percentage   composition   of  ethyl 
alcohol,  C0H5OH.  The  molecular  weight  is  24.00+  6.05 
+  16  =  46.05.      Let  x  =  %C,  y  =  %H  and  z  =  %0. 
24.    :  46.05  :  :  X  :  100.   x  =  52.117%  C. 
6.05  :  46.05  :  :  y :  100.    y  =  13- 13^%  H. 
16.       :  46.05  :  :  z  :  100.    z  =  34-745%  Q- 
To  prove     x  +  y  +  z  =  1 00  %. 
By  similar  proportions  the  percentage   composition 
of    any   substance    whose   formula    is    known   can  be 
calculated. 

CALCULATION    OF    PERCENTAGE    TO    THE    DRY    BASIS. 

Ores  and  other  material  often  contain  moisture:  it  is 
frequently  necessary  to  calculate  what  the  percentage  of 
a  constituent  would  be  if  this  water  were  removed  and 
the  reverse. 


20       CALCULATION  OF  PERCENTAGE  TO  THE  DRY  BASIS. 

If  an  ore  contains  io%  of  moisture  and  42%  of 
manganese,  what  will  be  the  percentage  of  manganese 
on  the  dry  basis  ?  Let  us  represent  the  weight  of  the  wet 
sample  by  100  parts;  then  after  drying  we  will  have 
100 — 10  (the  parts  of  water) :  let  x  =  the  percentage  of 
manganese  on  the  dry  basis;  then,  as  no  manganese  is  lost 
or  gained  by  the  removal  of  water,  we  have  42  X  100  =  x 
(100-10).  X  =  46.66%  or,  in  general  if  we  divide  the 
percentage  by  100  minus  the  percentage  of  moisture  we 
get  the  percentage  on  the  dry  basis. 

If  an  ore  contains  58%  of  iron  on  the  dry  basis,  what 
is  the  percentage  of  iron  when  18%  of  water  is  present? 

58  :  X  :  :  100  :  82.     x  =  47.56 
for   the  original  ore  may  be  regarded  as  made  up  of  82 
parts  of  dry  ore  containing  58%  of  iron  and  18  parts  of 
water  containing  no  iron. 

These  examples  are  of  very  great  importance  as  the 
quantity  of  moisture  in  a  shipment  of  ore,  coal,  etc.,  may 
be  quite  different  when  it  arrives  at  its  destination  from 
that  which  it  contained  when  shipped.  In  order  that  the 
chemists  for  the  buyer  and  seller  may  agree,  their  results 
must  be  calculated  to  a  common  basis,  usually  the  dry 
basis. 

If  a  carload  of  ore  were  sold  as  containing  42%  of 
lead,  and  the  buyer  found  40%  of  lead,  and  6%  of  water, 
what  must  have  been  the  percentage  of  water  in  the 
original  sample,  provided  both  chemists  were  perfectly 
accurate  ? 

The  buyer's  chemist  found  that  94  parts  of  dry  ore 
contained  40  parts  of  lead.  Let  x  =  number  of  parts  to 
contain  42  parts  of  lead ;  then 

4o:42::94:x.     x  =  98.70. 

The  original  ore  must  have  contained  1.30%  of  water. 


CALCULATION   OF   PERCENTAGES  TO  THE   DRY   BASIS.  21 

At  times  samples  are  received  in  a  wet  condition,  and 
as  the  water  will  be  constantly  evaporating  in  the  labora- 
tory, the  sample  must  be  weighed,  and  air-dried  at 
once. 

A  wet  sample  of  coal  weighed  800  grams  ;  this  was 
dried  at  a  temperature  of  60°  C.  and  lost  S%  of  water. 
The  partially  dried  sample  was  analyzed  and  found  to 
contain  2.5%  of  water.  What  was  the  percentage  of 
water  in  the  original  sample  ? 

The  air-dried  sample  was  made  up  of  97.5  parts  of 
coal  and  2.5  parts  of  water.  The  wet  sample  was  made 
up  of  92  parts  of  the  partially  dried  sample  and  8  parts 
of  water. 

To  get  the  water  in  92  parts  of  the  partially  dried 
sample,  we  have : 

100:  92  : :  2.5  :  X.     x  =  2.30. 

To  get  the  total  water  in  the  wet  sample  2.30  +  8 
=  10.30.  Ans.     10.30%. 

Let  us  suppose  that  the  analysis  of  the  air-dried  sample 
gave  the  following  results  :  Moisture  2.50%.  Volatile 
combustible  matter  18%.  Fixed  carbon  65.50%.  Sul- 
phur 2%.     Ash  1 2%.     What  are  the  percentages  on  the 

dry  basis  ? 

18     ^0.975  =  18.46%. 

65.5^0.975  =  67.18%. 

2     -^0.975=    2.05%. 

12     -^  0.975  =  12.31%. 

100. 

What  are  the  percentages  in  the  original  sample  ? 


22      CALCULATION   OF   PERCENTAGES   TO  THE   DRY   BASIS. 

i8     X  0.92  =  16.56%. 
65.5  X  0.92  =  60.26%. 

2       X0.92=      1.84%. 
12       XO.92  =   11.04%. 

Water  10.30%. 

100. 

These  results  can  also  be  obtained  by  multiplying  the 
weights  in  one  gram  on  the  dry  basis  by  100  minus  the 
percentage  of  water ;  in  this  case  by  89.70. 


EXAMPLES. 


Calculate  the  empirical  formulae  from  the  following 
analytical  results. 

1.  Lead  68.30%.  Sulphur  10.55%.   Oxygen  by  dif- 
ference, 21.15%.  Ans,     PbSO^. 

2.  Arsenic  41-32%.  Chlorine  58.66%. 

Ans.    AsClg. 

3.  Potassium  35.56%.  Iron  17.00%.  Cyanogen  by 
difference,  47-44%.  Ans.     KgFe  C,N^. 

4.  Carbon  37.5%.  Hydrogen  12.5%.  Oxygen  by 
difference,  50.%.  Ans,     CH4O. 

5.  Carbon  40.67%.  Hydrogen  8.47%.  Nitrogen 
^3'73%-    Oxygen  by  difference,   27.13%. 

Ans.     C2H5NO. 

6.  Carbon  46.75%-  Hydrogen  6.48%.  Nitrogen 
36.36%.  Oxygen  by  difference,    10.41%. 

Ans.     C6H10N4O. 
Calculate  the  formulae  of  the  following  minerals. 

7.  Pyrargyrite.  (*)  Sulphur  17.81%.  Antimony 
22.45%.     Silver  59.75%.  Ans.     AggSbSg. 

8.  Enargite.  Sulphur  32.69%.  Arsenic  19.47%. 
Copper  4  7. 84%.  A  ns.     Cug  ASS4. 

9.  Bournonite.  Sulphur  19.36%.  Antimony  23.57%. 
Arsenic  0.47%.  Lead  41.95%.  Copper  13.27%.  Iron 
0.68%.      Ans.     2PbS,CuoS,Sb2S3  or  3CPbCuo)S,SboS3. 

10.  Atacamite.  Chlorine  16.45%.  Copper  14.72%. 
Copper  oxide  (CuO)  55.26%.     Water  13.57%. 

Ans.     CuoCl  H3O3. 

*  These  percentages  are  taken  from  Dana's  Mineralogy. 

(23) 


24  EXAMPLES. 

11.  Barytocalcite.  Carbonic  acid  29.44%.  Barium 
oxide  50.36%.  Calcium  oxide  19.22%.  Manganous 
oxide  0.25%.  Ans.     BaC03,CaC03  or  RCO3. 

12.  Given  the  following  analysis  of  augite:  SiOo 
54.28%,  AL03  0.5i%,  Fe,03  0.98%,  FeO  i.9i%,Mgd 
17.30%,  CaO  25.04%.  Calculate  its  formula  and  oxy- 
gen ratio.  Ans.     CaMgSioOg  and  i  :  2. 

13.  Find  the  type  formula  and  oxygen  ratio  for  a 
mineral  whose  composition  is  Si02  38.03%,  AI0O3 
20.83%,  FeO  36.15%,  MnO  2.14%,  MgO  0.97%,  CaO 
2.73%.  Ans.     3RO,  R0O3,  3Si02  and  i  :  i. 

14.  The  mineral  Zirkelite  on  analysis  yielded  the 
following  results:  ZrOg  52.89%,  TiOo  14.95%,  ThOo 
7.31%,  CeAs  2.52%,  (Y2O3)?  0.21%,  U02i.4o%,  Fed 
7.72%,  CaO  10.79%,  MgO  0.22%,  loss  on  ignition 
1.02%.     What  is  the  formula?  (*) 

Ans.     RO.2  (Zr,  Ti,  Th,)02,. 
Calculate  the  percentage  composition  of  the  following 
compounds. 

15.  Potassium  Sulphate,  Kg  SO  4. 

Ans.     K,  44.91%.     S,  18.39%.     O,  36.70%. 

16.  Manganous  Sulphate,  MnS04. 

Ans.     Mn,  36.41%.     S,  21.22%.     O,  42.37%- 

17.  Lithium  Phosphate,  Li3(P04). 

Ans.     Li,  18.17%.     P,  26.70%.     O,  55.13%. 

18.  Zinc  Pyrophosphate,  Zn2P2  07. 

Ans.     Zn,  42.91%.     P,  20.34%.     O,  36.75%. 

19.  Urea,  CH4N2O  and  Ammonium  Cyanate,  NH4 
CNO. 

Ans.    0,19.96%.    H,  6.71%.    N,  46.71%.    0,26.61%. 

20.  Benzyl  Chloride,  CeHgCHgCl. 

Ans.     C,  66.40%.     H,  5.58%.     CI,  28.02. 

*  Mineralogical  Magazine,  1895-97,  XI,  180. 


EXAMPLES.  25 

21.  Nitrotoluene  C6H4(NOo)  CH3. 

Ans.  C,  61.27%.  H,  5.14%.  N,  10.24%.  O, 
23-34%. 

22.  Indigo  CieHioNsOg. 

Ans.    C,  73.24%.  H,  3.84%.  N,  10.71%.  O,  12.20%. 

23.  Potassium  ferrocyanide,  crystallized,  K4Fe(CN)6, 
3H2O.         Ans,    K,  37.04%.     Fe,  13.22%.     C,  17.03%. 

N,  19.93%.     H,  1.43%.     O,  1 1.35%. 

24.  A  sample  of  ore  contained  20%  of  lead,  14%  of 
zinc  and  14%  of  water.  What  are  the  percentages  of 
lead  and  zinc  on  the  dry  basis. 

Ans.     Lead  23.25%,  Zinc  16.28%. 

25.  A  clay  was  partially  dried  and  then  contained 
SiOo  50%,  H2O  7% ;  the  original  sample  contained 
1 2%  of  water.  What  is  the  percentage  of  Si02  in  the 
original  sample.  Ans,     47.31%. 

26.  A  sample  of  copper  ore  contained  18%  of  copper 
and  5%  of  water.  What  would  be  the  percentage  of 
copper  when  four-fifths  of  the  water  were  evaporated  ? 

Ans,     18.75%. 

27.  A  rich  silver  ore,  containing  3%  of  water  gave 
2100  ounces  per  ton  of  silver.  What  is  the  silver  in 
ounces  per  ton  on  the  dry  basis?  Ans.     2165  oz. 

28.  A  wet  sample  of  coal  gave  6%  of  water  when 
dried  at  70°  C.  This  partially  dried  sample  when  ana- 
lyzed showed  1.5%  water,  and  1.5%  sulphur.  What 
were  the  percentages  of  water  and  sulphur  in  the  original 
sample?  Ans.     7.41%  and  1.41%. 

29.  Ten  grams  of  a  silicate  mineral  lost  one  gram  of 
water  when  dried  at  100°  C.  Five  grams  of  the  sample 
dried  at  100°  C.  lost  o.i  gram  when  dried  at  250°  C. 
One  gram  of  the  sample  dried  at  250°  C.  lost  0.0 1  gram 
at  a  red  heat.  What  was  the  total  percentage  of  water 
in  the  silicate  ?  Ans.     1 2.682%. 


CHAPTER   III. 

CALCULATIONS    OF    MIXTURES    HAVING   A    COMMON 
CONSTITUENT. 

To  calculate  the  weights  of  chlorine  and  bromine  in 
a  mixture  of  silver  chloride  and  bromide. 

If  we  have  a  solution  containing  chlorides  and  bro- 
mides and  wish  to  determine  the  weights  of  chlorine  and 
bromine  present,  we  can  precipitate  these  elements  by 
silver  nitrate  and  weigh  the  resulting  mixture  of  AgCl 
and  AgBr,  weight  1.50  grams.  Then  by  scorification  or 
some  other  method,  determine  the  silver;  weight  i.oo 
gram.  We  have  the  atomic  weights  of  silver  107.93, 
chlorine  35.45,  bromine  79.96,  the  weight  of  the  mix- 
ture and  the  weight  of  the  common  constituent,  silver. 

If  all  the  silver  were  present  as  chloride  the   weight 
would  be  1.32845  grams:  for  as  we  know  the  formula 
and  atomic  weights  we  have  the  proportion 
107.93  ••   143-38    ••        I.     :    X. 
At  wt.  Ag :  Mol.  wt.  AgCl : :  wt.  Ag :  wt.  AgCl,  which 
gives  X  =  1.32845,  the  weight  of  silver  chloride  corre- 
sponding to  the  silver  found.     But  we  have  1.50  grams  of 
mixed   chloride    and   bromide,    hence    1.50- 1.32845   or 
0-^7155  gram  is  the  excess  in  weight  due  to  the  greater 
atomic  weight  of  bromine  and  therefore  proportional  to 
the  weight  of  bromine  present.     So  to  obtain  the  weight 
of  bromine  we  have  : 

79-96     -     35.45      :      79.96      ::      0.1715        ^  x. 
At.  wt.  Br  -  At.  wt.  CI :  At.  wt.  Br  : :  excess  in  wt. :  wt.  Br. 

(26) 


MIXTURES  HAVING  A  COMMON  CONSTITUENT.  27 

X  =  0.30809,  weight  Br.     The  weight  of  chlorine  will  be  : 
Weight  silver,  chlorine  and  bromine  1.50000  grams 

weight  silver  (i.)  and  bromine  (0.30809)   1.30809  grams 

or  0.19191  gram 

To  calculate  the  chlorine  directly  :  find  the  weight  of 
silver  bromide  corresponding  to  the  silver  present. 
107.93  :  187.89  ::  I  :  X.     x  =  1.74085  grams. 
But  we  have  only  1.5  grams  of  the  mixture  so  1.74085  — 
1.5  or  0.24085  is  the  deficiency  due  to  the  lower  atomic 
weight  of  chlorine  and  proportional  to  the  weight  of 
chlorine  present. 
Let  y  =  weight  of  chlorine,  then  we  have 

79-96  -  35.45  :  35.45  :  :  0.2408  :  x 
and  X  =  o.  191 79  the  weight  of  chlorine  present,  which 
confirms  the  result  obtained  by  difference. 

It  would  be  more  logical  to  say  that  the  excess  in 
weight  is  due  to  the  greater  molecular  weight  of  AgBr 
or  the  deficiency  due  to  the  lower  molecular  weight  of 
AgCl,  but  this  only  means  the  addition  of  107.93  in  both 
cases  which  does  not  affect  the  result  after  subtraction. 

Perhaps  the  following  method  is  more  easily  compre- 
hended from  a  mathematical  standpoint  but  it  is  longer 
and  the  use  of  logarithms  is  recommended. 

Let  X  =  weight  of  AgCl  and  y  =  weight  of  AgBr, 
then 

^^^'^^  X  =  wei2:ht  of  silver  in  AgCl  and 
143.38  ^  ^ 

^Z^'V"  y  =  weight  of  silver  in  AgBr  ;  so,  using  the  same 
187.89'^  ^ 

weights  as  before  we  have  the  two  equations, 

(i)  x  +  y  =  1.50  and 

^'^43.38     +187.89^ 


28  MIXTURES  HAVING  A  COMMON  CONSTITUENT. 

reducing  to  decimals  we  have  0.75275  x-f-  0.57443  y  =  i 
substituting  for  x  its  value  in  terms  of  y  from  equation 
(i)  we  have 

0.75275  (i-5  -  y)  +  0.57443  y  =  I     Solving 
0.1291  =  0.1783  y.     y  =  0.7241,  wt.  AgBr. 

substituting  this  value  in  equation  (i)  we  get  x  =  0.7759 
wt.  AgCl ,  from  which  by  the  following  proportions, 

143-38  :  35.45  '  •  0.7759  :  wt.  CI  and 
187.89  :  79.96  :  :  0.7241  :  wt.  Br ; 

we  get  wt.  Br  0.30815  and  wt.  CI  o.  191 84  gram. 

Suppose  we  have  the  weight  of  a  mixture  of  potas- 
sium and  sodium  sulphates,  0.3710  gram,  and  the  weight 
of  SO 3,  0.200  gram  ;  to  calculate  the  weights  of  K2O 
and  NagO.     Calculate  the  SO3  all  to  Na2S04, 

80.06  :  142.16  : :  0.2  :  x.     x  =  0.3551. 

but  as  we  have  0.3710  gram  of  mixed  sulphates,  the 
excess  in  weight  0.0159  gram  is  due  to  the  excess  in 
molecular  weight  of  K2O  over  NaaO  and  is  therefore 
proportional  to  the  weight  of  KgO,  so  we  have  the  pro- 
portion, 

32.2  :  94.3  : :  0.0159  :  x.     x  =  0.0466 
K2O  0.0466  gram.  K2O.  -|-Na20-|-S03  =0.3710 gram, 
SO 3  0.2000  gram.  KgO.  +  SO  3  =  0.2466  gram 

0.2466  gram.  Na2  0  =0.1 244  gram 

To  solve  this  example  by  the  other  method,  let  x  = 
weight  of  K2SO4  and  y  =  weight  of  Na2S04,  then  x  +  y 
=  0.371. 

The  molecular  weight  of  SO  3  is  80.06  and  that  of 
K2SO4  is  174.36;  therefore 

-— ^ — -  X  represents  the  weight  of  SO 3  present  as  K2SO4 


MIXTURES  CONTAINING  A  COMMON   CONSTITUENT.  29 

and  similarly  — '- — -  y  the  weight  of  SO  3   present  as 

Na2S04,  SO  we  have  for  a  second  equation, 

80.06        ,    80.06  ^  1.1.  1 

X  H y  =  0.2  or  alter  multiplyingf  and  re- 

174.36       '    142.16  ^  f  J    s 

moving  unnecessary  fractions, 

1 1381.3  X  + 13959.3  y  =  4957.4 

substituting  from  first  equation  x  =  0.371  —  y  we  have 
11381.3  (0.371  -  y)  +  13959.3  y  =  4957.4  or 
2578.0  y  =  734.9;  y  =  0.2851,  weight  of  Na2S04 
and  as  x  +  y  =  0.3710.  x  =  0.0859,  weight  of  K2SO4. 
From   which    the    following   proportions   will   give   the 
weights  of  K2O  and  Na2  0 

94.3  :  174.36  : :  wt.  K2O  :  0.0859.  wt.  K2O  =  0.0464. 
62.1  :  142.16  : :  wt.  Na20  :  0.2851.  wt.  Na20  =  0.1245. 
The  following  is  a  more  practical  example.  One 
gram  of  commercial  sodium  bicarbonate  gave  CO  2, 
0.5097  gram  and  Na2  0,  0.3681  gram.  Calculate  the 
weights  of  NaHCOg  and  Na2C03.  Let  x  be  the  weight 
of  NagCOs  and  y  the  weight  NaHC03  ;  then 

(i)       ^"^    X  +  w^^  y  =  0.5097  and 
^  ^     106. 1      ^84.06  ^  ^  ^^ 

/  X      62.1        ,     62.1  ^o 

(2)     — 7— x  +  ^^ — y  =  0.3681. 
^  ^     106.1  168.12  ^ 

The  molecular  weight  of  NaHCOg  is  doubled  in  the 
second  equation  as  it  takes  two  molecules  to  give  one 
Na2  0.  Reduce  the  fractional  coefificients  to  the  nearest 
decimals, 

(i)     0.4147  x  + 0.5234  y  =  0.5097 
(2)     0.5853  X  +  0.3694  y  =  0.3681 


30  MIXTURES  CONTAINING  A  COMMON  CONSTITUENT. 

substituting  for  x  in  the  second  equation  its  value  in 
terms  of  y  we  get 

-5853  ^°-'°''-,°,r''^  +  °-3^94  y  =  0.3681 
0.7192  -  0.7385  y  +  0.3694  y  =  0.3681 

0.3691  y  =  0.351 1,     y  =  0.9512,  weight  of  NaHCOg. 

Substituting  for  y  in  equation  (i)  0.9512  we  get 
0.4147  x  +  0.5234  (0.9512)  =  0.5097  which  gives  x  = 
0.0285,  weight  of  NagCOg. 

This  problem  may  also  be  done  as  follows  :  Calculate 
all  the  CO 2  to  NaHCOg,  according  to  the  proportion. 

44.  :  84.06  : :  0.5097  :  wt.  NaHCOg  : 
this  gives  0.9737  gram  ;  next  calculate  the  weight  of 
NagO  necessary  to  combine  with  this  to  form  the  0.9737 
gram  of  NaHCOg.  This  is  0.35967  gram,  but  we  have 
present  0.3681  gram,  which  leaves  an  excess  of  0.00843 
gram.  This  must  be  employed  to  make  as  many 
NaHCOg  molecules  into  Na2COg  molecules  as  possible 
and,  as  one  NagO  furnishes  the  sodium  oxide  necessary 
to  convert  2 NaHCOg  into  2Na2COg,  to  find  the  amount 
of  Nag  CO  3  required  by  the  excess  of  Na2  0  we  have  :  — 
62.1  :  212.2  :: 

Mol.  wt.  NagO  :  2  (Mol.  wt.  Na2COg)  : : 

0.00843  •         ^ 

wt.  Na2  0  in  excess  :  w^t.  NagCOs 
x  =  0.0288,  weight  Na2COg. 

The  weight  of  Na2  0  present  as  Na2COg  is  twice  the 
excess  found  or  0.01686  gram.  The  rest  of  the  NagO, 
0.3681  —0.01686  or  0.35 1 24 gram  is  present  as  NaHCOg. 
To  find  this  weight :  — 

0.35124  :  X  ::  62.1  :  168.12.  x  =  0.9509,  weight  of 
NaHCOg. 


FORMATION  OF  MIXTURES  OF  DEFINITE  COMPOSITION.    31 

These  examples  show  the  results  to  be  the  same  by 
either  method,  the  figures  should  be  carried  out  to  four 
decimals  or  at  least  three  significant  figures  so  that  no 
appreciable  error  in  calculation  shall  be  added  to  the  un- 
avoidable errors  of  analysis. 

FORMATION     OF    MIXTURES     OF     DEFINITE    COMPOSITION. 

Suppose  we  wish  to  form  loo  lbs.  of  a  mixture  con- 
taining io%  of  calcium  carbonate  and  we  have  at  our 
disposal  lots  containing  7%  and  23%  respectively. 
How  many  pounds  of  each  must  be  taken  ? 

One  lot  contains  3%  less  than  the  desired  mixture, 
the  other  13%  more,  so  if  we  take  3  parts  of  the  richer 
to  13  of  the  poorer  we  shall  have  a  mixture  containing 
10%. 

3  :  16  :  :  X  :  100.     x  =  18.75  lbs.  (23%). 
13  :  16  :  :  y  :  100.     y  =  81.25  lbs.  (7%), 

When  the  proportions  cannot  be  seen  by  inspection 
the  following  method  can  be  used  : 

Given  two  iron  ores,  one  containing  0.42%  of  phos- 
phorus and  the  other  o.  i  y%.  How  much  of  each  must 
be  mixed  to  give  ten  tons  containing  o.  20%  of  phos- 
phorus? Let  X  be  the  weight  in  tons  of  the  0.42%  ore 
and  10  —  X  the  weight  of  the  0.17%  ore;  then, 

X  (0.0042)  +  (10 — x)  0.0017  =  10  (0.002). 

This  equation  is  based  on  the  equality  in  weight  of  phos- 
phorus in  the  sum  of  the  two  portions,  and  in  the  mixture. 

Solving,  we  get  x  =  1.2  tons,  10 — x  =8.8  tons. 

Given  waste  mixed  acid  from  nitrating  whose  com- 
position is  H0SO4  62.178%,  HNO,  19.066%.  HoO 
18.756%;  sulphuric  acid  containing  97%  H0SO4;  nitric 
acid  containing  87%  HNO3.    What  weights  of  each  must 


32    FORMATION  OF  MIXTURES  OF  DEFINITE  COMPOSITION. 

be  taken  to  give  looo  lbs.  of  a  mixture  containing 
6i%  HoS04,22%  HNO3  and  iy%  H^O  without  adding 
water  ? 

Let  X  be  the  weight  of  waste  acid, 
y         "         "        ''  sulphuric  acid, 
z  "         "        '*  nitric  acid, 

then  X  (0.6218)  is  the  weight  of  H2SO4  in  the  waste 
acid,  y  (0.97)  the  weight  of  H2SO4  in  the  sulphuric  acid, 
z  (0.13)  the  weight  of  water  in  the  nitric  acid  and  so  on. 
1000  lbs.  of  the  desired  mixture  must  contain  610  lbs. 
H2SO4,  220  lbs.  HNO3  and  170  lbs.  of  water.  We  have 
therefore  the  Following  equations  based  on  equality  of 
weight: 

(i)  X  (0.6218)  +y  (0.97)  =  610. 

(2)  X  (0.19066)  +  z  (0.87)  =  220. 

(3)  X  (0.18756)  +  y  (0.03)  +  z  (0.13)  =  170. 

610 X     (0.6218)  ^00  /      /:       \ 

y  = ^^ ^  =  628.8  — X  (0.641) 

0.97 

220 — X    (0.19066)  ^     ^  /■  \ 

Z   =  ^^ ^ ^    =    2^2.7 X  (0.2  191) 

0.87  ^   ^      ^      ^  ^ 

substituting  in  equation  (3) 

0.18756  X  +  18.864 — 0.01923  x  +  32.851 — 0.02848  X 
=  170. 

0.13985  X  =  118.285.     ^  =  845.68. 
substituting  in  equation  (i) 

^  610  — (845-68x0.6218).    ^  ^  3^^^^^ 

0.97 
substituting  in  equation  (2) 

z  =  220— (845.68x0.19066).     ^  _  5.  - . 
0.87  ^'^^' 

Proof  845.68  +  86.77  +  67.54  =  999.99. 


EXAMPLES. 

1.  Given  a  mixture  of  KCl  and  NaCl  :  weight  1.331 
grams  ;  and  weight  of  chlorine  0.709  gram.  What  ar*^ 
the  weights  of  K  and  Na  ? 

Ans.     K  0.3915  gram.     Na  0.2305  gram. 

2.  Given  a  mixture  of  PbS04  and  BaS04  5  weight  4 
grams;  and  weight  of  SO 4  1.62  grams.  What  are  the 
weights  of  PbS04  and  BaS04  ? 

Ans,     PbS04  0.2738  gram.     BaS04  3.7262  grams. 

3.  Given  a  mixture  of  HgClg  and  HgCl  weight  2.35 
grams  and  chlorine  0.45  gram.  What  is  the  weight  of 
each  chloride  ? 

Ans.     HgCl  1.4847  grams.     HgCls  0.8653  gram. 

4.  Given  a  mixture  of  Agl  and  AgBr;  weight  4.22 
grams  and  weight  of  silver  2. 11  grams.  What  are  the 
weights  of  iodine  and  bromine  ? 

Ans     Iodine  1.4792  grams.     Bromine  0.6308  gram. 

5.  Given  a  mixture  of  NaHS04  and  KHSO4  • 
weight  0.5  gram  and  weight  SO4  0.395  gram.  What 
are  the  weights  of  K  and  Na  ? 

Ans.     K  0.0148  gram.     Na  0.0860  gram. 

6.  Given  a  mixture  of  Kg  CO  3  and  KHCO3  ;  weight 
I  gram  and  weight  CO 2  0.40  gram.  What  is  the  weight 
of  each  carbonate  ? 

Ans.     K2CO3  0.3245  gram.     KHCO3  0.6755  gram. 

7.  Givenamixtureof  K2C4H4O6  and  KHC4H4O6  : 
weight  I  gram  and  weight  of  K  0.30  gram.  What  is  the 
weight  of  each  tartrate  ? 

(33) 


34  EXAMPLES. 

Ans.    K2C4H4O6  0.6668  gram.     KHC4H4O6  0.3332 
gram. 

8.  Given  a  sample  of  pure  dolomite,  MgCOg  and 
CaCOg  :  weight  i  gram  and  CO 2  weight  0.48  gram. 
Find  the  weights  of  CaO  and  MgO. 

Ans.     CaO  0.2841  gram.     MgO  0.2359  gram. 

9.  Given  a  mixture  of  CaH4  (P04)2  and  Cag  (P04)2 
weight  I  gram  and  PO4  weight  0.70  gram.     Calculate 
the  weight  of  each  phosphate. 

Ans  \  CaH4(P04)2  0.4402  gram. 
•  (  Ca3(P04)2  0.5598  gram. 

10.  Given  a  sample  of  commercial  potassium  bicar- 
bonate containing  0.4679  gram  of  KgO  and  0.4338  gram 
of  CO2  in  one  gram.  How  much  KHCO3  and  K2CO3 
is  there  present  ? 

Ans,     KHCO3  0.981 1  gram.     Kg  CO  3  0.0088  gram. 

11.  One  gram  of  manganese  ore  gave  on  analysis, 
manganese  0.2  5  gram  and  available  oxygen  0.05  gram. 
How  much  MngOg   and   MnOo  is  there  present?     By 
available  oxygen  is  meant  the  amount  over  that  neces- 
sary to  form  MnO. 

Let  X  =  wt.  MnoOa  and  y  =  wt.  MnOg  then  we  have 

I^x  +  |l:2?y  =  o.25.         Ans.     MuoOa  0.2245. 
157.98  86-99^ 

^^      x+^y  =  o.05.  Mn02    0.1482. 


157.98  86.99 

12.  Given  a  sample  of  argol  (crude  potassium 
bitartrate)  ;  one  gram  of  which  gave  0.2336  gram  of 
K2O  and  0.7310  gram  H2C4H4O6.  Supposing  that  there 
is  no  other  tartrate  or  potassium  salt  present,  how 
much  KHC4H4O6  and  K0C4H4O6  is  there  ? 

Let  X  =  wt.  K2C4H466  and  y  =  KHC4H4O6,  then 
we  have 


EXAMPLES.  36 

94.22.        ,        47.11  ^         J      I  SO.OQ        ,       I  SO.OQ 

-^ —  X  +    -If- —   y  =  0.2336  and     ^^ ^x4-      ^       < 

226.29  188.19^  ^^  226.29  188.19 

V  -  o  7^1  yln,   ^0.9013    gram  KHC4H4O6 

^       ^^'  ''I0.01S7   gram  K.QH^Og 

13.  Given    manganese    ores    containing    47%    and 

23%  of  manganese  respectively.     How   many  pounds 

of  each  must  be  mixed  to  give  a  ton  containing  39%  of 

y,       (    666.67  lbs.  2^9^  ore. 
mang^anese.  Ans.  <  '    1,        *^<0/ 

^  <  -^m-ZZ   lbs.  47%  ore. 

14.  Suppose  we  have  ten  tons  of  coal  containing 
3.5%  of  sulphur.  We  have  also  supplies  of  coal  con- 
taining 1.1%  and  1.6%  of  sulphur.  How  much  of  each 
shall  we  add  to  the  ten  tons  to  make  20  tons  containing 
2.5%  of  sulphur?     Ans.     2  tons  1.1%  and  8  tons  1.6%. 

15.  Given  iron  ores  containing  54%  of  iron  with 
0.60%  of  phosphorus  and  4  7%  of  iron  with  o.  20%  of  phos- 
phorus. What  is  the  percentage  of  iron  in  a  mixture  of 
these  containing  0.28%  of  phosphorus?     Ans,     48.4%. 

16.  Given  two  alloys:  a  brass  containing  copper 
65%,  zinc  35%  and  a  German  silver  containing  copper 
56%,  zinc  24%,  nickel  20%.  What  is  the  composition 
of  the  alloy  formed  by  melting  these  together  in  such  pro- 
portion that  it  contains  zinc  30%  ?  (Supposing  that  there 
is  no  loss  by  oxidation  or  volatilization  during  fusion). 

Ans.     Copper  60.90%,  zinc  30%,  nickel  9.09%. 

17.  Suppose  we  wish  to  form  100  lbs.  of  an  alloy 
containing  copper  70%,  zinc  20%,  and  tin  10%  and  we 
have  besides  pure  copper,  a  brass,  copper  %',  zinc  Yi  and 
solder,  zinc  ^,  tin  >^.    How  much  of  each  must  be  taken? 

Ans,     Solder  20  lbs.,  brass  30  lbs.,  copper  50  lbs. 

1 8.  Given  pure  silver  and  an  alloy  containing  890  parts 
of  silver  and  no  parts  of  copper.  How  many  ounces 
of  each  must  be  taken  to  form  1000  ounces  of  an  alloy, 
925  parts  of  silver  and  75  parts  of  copper? 

Ans,     681.8  ozs.  of  alloy,  318.2  ozs.  of  silver. 


CHAPTER  IV. 

CALCULATIONS  FROM  EQUATIONS. 

Chemical  reactions  are  best  represented  by  equa- 
tions ;  these  are  the  expression  of  definite  and  exact 
quantitative  changes  and  therefore  are  the  basis  of  most 
analytical  calculations. 

Take  for  example  the  simple  reaction 

BaClo  +  H2SO4  =  BaS04  +  2HCI, 
from  a  qualitative  standpoint  this  means  that,  when  bar- 
ium chloride  is  added  to  sulphuric  acid  or  the  reverse, 
barium  sulphate  and  hydrochloric  acid  are  formed.  This 
is  not  all  that  is  shown  by  the  reaction.  It  shows  that 
one  molecule  of  barium  chloride  acting  on  one  molecule 
of  sulphuric  acid  forms  one  molecule  of  barium  sulphate 
and  two  molecules  of  hydrochloric  acid.  Now  as  the 
relative  weights  of  the  different  atoms  composing  the 
molecules  are  known,  based  on  oxygen  =16,  we  can 
by  adding  these  together  get  the  relative  weights  of  the 
molecules ;  so  the  reaction  becomes, 

BaCl2  +  H2SO4  =  BaS04  +  2HCL 
137.4  +  2(35.45)  +  2(1.008)  +  32.06  +  64  = 

137.4+32.06  +  64  + 
2(35.45  +  1.008)  or  208.3+98.076  =  233.46  +  72.916  ; 

which  expresses  a  definite  relation  by  weight.  We  do 
not  know  the  absolute  weight  of  a  molecule  but  we  do 
know  very  accurately  their  relative  weights  ;  so  that  these 
proportions  are  true  when  expressed  in  parts  by  weight ; 
and,  whatever  the  unit  of  the  system  may  be,  grains, 

(36) 


CALCULATIONS  FROM  EQUATIONS.  37 

grams,  pounds,  or  tons,  we  shall  always  have  233.46  parts 
of  barium  sulphate  and  72.916  parts  of  hydrochloric  acid 
formed  by  the  action  of  208.3  parts  of  barium  chloride  on 
98.076  parts  of  sulphuric  acid. 

Besides  showing  the  products  of  the  reaction  and  the 
relations  by  weight,  the  equation  also  represents  relations 
of  energy  and  heat  absorbed  or  evolved.  These  will  be 
discussed  so  far  as  they  relate  to  analytical  calculations 
in  subsequent  chapters. 

Suppose  the  percentage  of  SO 3  is  desired  in  magne- 
sium sulphate.  The  salt  is  dissolved,  barium  chloride 
added  and  the  resulting  barium  sulphate  weighed.  Ac- 
cording to  the  preceding  reaction,  for  every  molecule  of 
SO  3  present  there  will  be  a  molecule  of  BaS04  precipi- 
tated ;  hence  to  find  the  weight  of  SO  3,  we  have  :  molec- 
ular weight  of  BaS04  :  molecular  weight  of  SO  3  :  : 
actual  weight  of  BaS04  :  corresponding  weight  of  SO 3, or 

233.46  :  80.06  : :  wt.  BaS04  •  ^t.  SO3 

weig^ht  BaS04  X  80.06  .  ,     c,^ 

or  2 ± =  weig^ht  SO  3 

233.46  ^ 

and  to  get  percentage  we  have  : 

weight  taken  for  analysis  (magnesium  sulphate)  :  100 :  : 
weight  SO3  :  x 

^^         weight  SO3X  100  ^       rentage  of  SO3.  G) 
weight  taken 
In  the  same  way  this  precipitate  of  BaS04  can  be 
used  to  determine  the  percentage  of  barium  or  barium 
oxide,  here  we  have  for  barium:  molecular  weight  of 


(*)  When  the  magnesium  sulphate  is  dissolved  in  water  it  is  partly  dissociated 
into  Mg  and  SO4  ions  which  react  with  the  Ba  and  CI  ions,  so  the  statement  "mole- 
cule of  SO3  present  in  solution  "  is  not  literally  correct  but  should  be  SO4  ions 
corresponding  to  molecules  of  SOj.  This  does  not  however  affect  the  truth  of  the 
proportions  given. 


38  CALCULATIONS  FROM  EQUATIONS. 

BaS04:  atomic  weight  Ba: :  weight  BaS04  -  weight  Ba, 
and  for  barium  oxide : 

mol.  wt.  BaS04  :  mol.  wt.  BaO  :  :  wt.  BaS04  :  wt.  BaO. 
or  233.46  :  153.4  :  :  wt.  BaS04  :  x. 

and  T — . — —    X  100  =  Percentage  BaO. 

wt.  taken  lor  analysis 

If  in  this  same  precipitate  the  sulphur  is  desired  (for 

example  in  coal  analysis)  we  have 

233.46  :  32.06  : :  wt.  BaS04  :  x  (wt.  S.) 

The  calculation  for  lead  sulphate  will  be  similar, 
though  the  proportion  will  be  different  on  account  of  the 
higher  atomic  weight  of  lead.  To  calculate  the  weight  of 
lead  corresponding  to  one  gram  of  PbS04  we  have : 

302.96  (Mol.  wt.  PbS04)  :  206.9  (At.  wt.  Pb)  : :  i  :  x. 

X  =  0.6829. 

Suppose  we  precipitate  MgNH4P04.  6H2O,  filter, 
wash,  ignite,  and  weigh.  On  ignition  this  is  converted 
into  magnesium  pyrophosphate,  MggPgOj,  which  can  be 
used  either  for  the  determination  of  magnesium  oxide 
or  phosphoric  anhydride.     We  have  the  reactions : 

MgClg  +  NH4OH  +  NaoHP04  =  MgNH4P04  + 
2NaCl  +  ~H20. 
and,  2MgNH4P04  +  heat  =  Mg2F20,+  2NH3+  HoO. 

For  every  molecule  of  the  ignited  precipitate  there 
must  have  been  present  in  the  original  solution  two 
atoms  of  magnesium,  two  atoms  of  phosphorus  or  one 
molecule  of  P2O5;  so  to  get  the  corresponding  amount 
of  MgO  we  have : 

Mol.  wt.  MggPsO; :  2  (Mol.  wt.  MgO)  : :  wt.  ppt.  :  x. 
or      222.72        :  80.72  ::  wt.  ppt.  :  x. 

and    ■  X  100  =  Percentage  of  MgO. 

wt.  taken 


CALCULATIONS  FROM  EQUATIONS.  39 

Similarly  to  get  the  corresponding  weights  of  phosphorus 
or  phosphoric  anhydride  we  have : 

Mg2P207  :  2P  : :  wt.  ppt.  :  x. 

222.72  :  62.         : :  wt.  ppt.  :  x. 

and  Mg2P207  :  P2O5      : :  wt.  ppt.  :  y. 

222.72  :  142.      : :  wt.  ppt.  :  y. 

The  calculations  for  the  pyrophosphates  of  zinc  and 
of  manganese  are  similar,  using  the  atomic  weights  of 
these  metals  instead  of  that  of  magnesium. 

Zinc  is  often  weighed  as  ZnNH4P04  :  in  this  case  for 
each  molecule  of  the  precipitate  there  will  be  but  one 
zinc  and  the  proportion  will  be 

178.47  :  65.4  : :  wt.  ppt.  :  x. 

All  direct  stoichiometrical  calculations  are  made  in 
this  way,  using  the  proper  atomic  weights  and  bear- 
ing in  mind  to  place  in  the  proportion  the  weight  of  the 
number  of  atoms  or  molecules  of  the  constituent  sought, 
which  the  precipitate  or  the  residue  weighed  contains. 

Indirect  Determinations. — In  the  material  weighed 
the  element  to  be  determined  may  not  be  present  at  all. 

For  example  :  Potassium  is  .precipitated  quantita- 
tively as  KgPtCle :  this  may  be  weighed  as  such  or  may 
be  ignited  with  a  little  oxalic  acid,  the  soluble  potassium 
chloride  washed  out,  and  the  spongy  platinum  weighed. 
The  reactions  are 

2KCI+  HsPtCle  =  KoPtCl6+  2HCI. 

To  calculate  the  weight  of  KoO  from  the  weight  of  the 
double  chloride  we  have 

Mol.  wt.  KoPtCle :  Mol.  wt.  KoO  :  :  wt.  ppt.  :  x. 
485.8  :         94.3  :  :  wt.  ppt.  :  x. 

If  platinum  is  weighed : — each  molecule  of  KoPtCl^ 
has  one  atom  of  platinum  and  two  of  potassium,  so  for 


40  QUANTITY  OF  REAGENTS. 

each  atom  of  platinum  weighed  there  must  have  been 
two  atoms  of  potassium,  equivalent  to  a  molecule  of 
KoO  in  the  solution  to  cause  its  precipitation,  hence  we 
have 

At.  wt.  Pt  :  Mol.  wt.  K2O  :  :  wt.  residue  :  x 
or  194.8         :  94.3  :  :  wt.  residue  :  x 

•V- 

and  in  either  case — X  100  =  Percent  of  KgO. 

wt.  taken 

It  must  always  be  kept  in  mind  that,  when  there  is  a 
change  in  composition  during  ignition,  it  is  the  formula 
of  what  is  actually  weighed  which  gives  the  basis  for 
calculation.  For  example,  cobalt  is  often  separated 
from  nickel  by  precipitation  as  2C0  (N0o)3,  6KNO0  and 
after  filtering,  etc.,  is  converted  into  C0SO4  and  K2SO4 
by  sulphuric  acid.  The  properties  of  the  double  nitrite 
are  studied  for  the  purpose  of  separation  but  all  calcu- 
lations are  based  on  the  sulphates  weighed.  To  illus- 
trate :  One  gram  of  nickel  matte  was  taken  for  analysis 
and  the  mixed  sulphates  of  cobalt  and  potassium  weighed 
0.0896  gram.  What  is  the  percentage  of  cobalt  ?  We 
have  the  reaction, 

2C0  (N0o)3,  6KNO2+  5H2SO4  =  2C0SO4+  3K2SO4 
+  5H2O+7NO2+5NO. 
so  for  every  2C0SO4+  3K0SO4  there  must  be  2  Co  or 

833.2  :  118.  ::  0.0896  :  x.  x  =  0.0127  gram  or 
1.27%  Co. 

QUANTITY  OF  REAGENTS. 

In  analytical  as  well  as  technical  work  it  is  often  ne- 
cessary to  know  what  quantity  of  a  reagent  to  add  to 
cause  a  reaction. 

For  example.  How  much  4%  ammonium  oxalate 
is  necessary  to  precipitate  as  calcium  oxalate  the  lime  in 
one  gram  of  limestone  ?    We  have  the  reactions  ; 


QUANTITY  OF  REAGENTS.  41 

CaCOa  +  2HCI  =  CaCL  +  HoO  +  CO2 
CaCL  +  (NH4)  C2O4  =  CaCoO^  +  2NH4CI. 

We  see  by  comparing  the  two  equations  that  one  mole- 
cule of  CaCOg  requires  one  of  (NH4)2C204.  So,  assum- 
ing that  the  limestone  is  pure  calcium  carbonate,  we 
have;  Mol.  wt.  CaCOg  :  Mol.  wt.  (NH4)X204  :  :  wt. 
limestone  :  wt.  ammonium  oxalate  or 

100. 1  :  124. 14  : :  I  :  X.     x  =  1.2402  grams, 

but  a  4%  (by  weight)  solution  is  to  be  used ;  to  find 
what  weight  of  solution  contains  1.2402  grams  of  the  salt 
we  have 

1.2402  :  x  : :  4  :  100.     x  =  31.005  grams. 

It  is  not  to  be  inferred  that  this  is  necessarily  the 
best  amount  to  add  but  it  is  certain  that  less  will  be  in- 
sufficient for  complete  precipitation. 

In  the  precipitation  of  zinc  as  zinc  ammonium  phos- 
phate, it  is  considered  advisable  to  add  three  times  the 
amount  of  the  precipitant  actually  required.  How  many 
grams  of  a  15%  solution  of  NH4NaHP04  should  be 
added  to  a  solution  containing  0.25  gram  of  zinc?  The 
reaction  is : 

Zn(N03)o+  NH4NaHP04+ NH4OH.  = 

ZnNH4P04  +  NH4NO3  +  NaNOa  +  HoO 

or  one  molecule  of  precipitant  for  one  atom  of  zinc : 
Hence  to  satisfy  the  equation  0.5245  gram  are  required 
for, 

137-13  •  65.4  : :  X  :  0.25.     X  =  0.5242. 
The  corresponding  weight  of  a  15%  solution  is  3.496 
grams,  for 

0.5242  :  X  : :  15  :  100.     x  =  3.495. 
This  is  the  weight  of  a  1 5%  solution  necessary  to  satisfy 


42     TO  DETERMINE  WHICH  REAGENT  IS  MOST  ECONOMICAL. 

the  equation  but  we  are  to  add  three  times  this  amount 
or  10.485  grams. 

To  obtain  a  15%  solution  of  the  active  constituent: 
Microcosmic  salt  contains  four  molecules  of  water  of 
crystallization,  so  to  make — say  100  grams  of  a  15% 
solution  we  must  first  find  what  weight  of  the  crystallized 
salt  is  required  to  give  15  grams  of  the  active  con- 
stituent. 

Mol.  wt.  NaNH4HP04  :  Mol.  wt.  NaNH4HP04,  4H2O  :  : 
15  :  X  or  137.13  :  209.19  :  :  15  :  x.     x  =  22.882. 

So  we  must  weigh  out  22.882  grams  and  dissolve  it  in 
100  —  22.882  grams  or  77.1 18  grams  of  water. 

TO    DETERMINE    WHICH    REAGENT    IS    MOST    ECONOMICAL. 

Cases  often  arise  where  a  reaction  can  be  performed 
by  several  reagents,  so  that  it  is  of  importance  to  use 
the  most  economical,  not  necessarily  the  lowest  priced. 
For  example,  which  is  more  economical  as  an  oxidizing 
agent,  potassium  chlorate  at  8  cents  a  pound  or  sodium 
chlorate  at  10  cents? 

One  molecule  of  either  will  yield  three  atoms  of  oxy- 
gen, so  the  oxygen  in  a  pound  of  KCIO3  is  — ^ — >  lb.  and 

122.6 

the  oxygen  in  a  pound  of  NaClOg  is— ^^ —  lb.     The  most 

106.5 

economical  is  the  one  which  gives  the  most  oxygen  for 

one  cent.     To  compare  these  we  have, 

KCIO3  — ^ —  -^  8  or  0.04894  lb.  for  one  cent. 
122.6 

A  R 

NaClOa  -%—  -^  10  or  0.04507  lb.  for  one  cent. 
106.5 

Which  shows  potassium  chlorate  to  be  more  economical 

at  the  prices  given. 


EXAMPLES. 


1.  What  is  the  weight  of  BaO  in  1.9327  grams  of 
BaCr04  ?  Ans.     1.1695. 

2.  What  is  the  weight  of  CaO  in  2.9478  grams  of 
CaS04  ?  Ans.     1.2145. 

3.  What  weight  of  MnCOg  yields  on  ignition  1.450 
grams  of  Mn304  ?  Ans.     2.1845. 

4.  How  much  arsenic  is  there  in  5  grams  of  AsgSg  ? 
ofAsgSs?    ofMgaAssO^? 

Ans.     (a)  3.0465,  (b)  2.4170,  (c)  2.4135. 

5.  How  much  Agl  can  be  made  from  a  pound  of  pure 
silver  ?     How  much  AgBr  ?     How  much  AgCl  ? 

Ans.     2.1753  lbs.  Agl,  1.7409  lbs.  AgBr,  1.3285  lbs. 
AgCl. 

6.  1. 10  grams  of  stibnite  gave  on  analysis  0.5987 
gram  of  Sb2  04.     What  is  the  percentage  of  antimony? 

Ans.     42.98%. 

7.  How  many  pounds  of  water  will  it  take  to  slack 
100  lbs.  of  quicklime  (CaO)?  Afis.     32.11  lbs. 

8.  How  many  pounds  of  red  lead  (Pb3  04)  can  be 
made  from  a  ton  of  litharge  (PbO)  ? 

Ans.     2048  lbs. 

9.  How  many  grams  of  MnOg  and  of  H2SO4  are  re- 
quired to  give  10  grams  of  oxygen? 

Ans.     61.297  grams  H2SO4,  54-375  grams  MnOg. 

10.  One  gram  of  coal  gave  0.2634  gram  of  BaS04 
by  Eschka's  method  ;   i  gram  of  MgO  and  0.5  gram  of 

(43) 


44  EXAMPLES. 

Na2C03  were  used.  It  was  found  that  both  contained 
sulphur:  lo  grams  of  MgO  and  5  grams  of  NaoCOg 
(together)  gave  0.1654  gram  of  BaS04.  What  is  the 
percentage  of  sulphur  in  the  coal?         Ans.      3.39%. 

1 1.  How  many  pounds  of  20%  H0SO4  will  it  take  to 
neutralize  a  ton  of  xNaXOg?    of  NaHCOg?  of  CaCOa? 

A71S,     (a)  92437  lbs.,  (b)  5833.8  lbs.,  (c)  9797.B  lbs. 

12.  How  many  pounds  of  10%  HCl  will  it  take  to 
neutralize  a  ton  of  Na2 CO 3?  ofNaHCOs?  ofCaCOe? 

Ans.     (a)  13745  lbs.,  (b)  8674  lbs.,  (c)  14569  lbs. 

13.  Suppose  one  gram  of  silver  is  dissolved  in  nitric 
acid  and  to  it  is  added  0.25  gram  of  pure  dry  sodium 
chloride.  What  percentage  of  the  silver  remains  in  so- 
lution? Ans.      53.88%. 

14.  Suppose  0.25  gram  of  sodium  bromide  were 
added  to  a  solution  of  one  gram  of  silver.  What  per- 
centage of  silver  remains  in  solution  ? 

Ans.     73.81%. 

15.  A  dolomite  contains  98%  of  calcium  and  mag^ 
nesium  carbonates,  2%  of  SiOg  and  10%  of  MgO. 
What  is  the  percentage  of  COo?  Ans.     44.79%. 

16.  One  gram  of  ore  containing  nickel  and  cobalt 
gave  0.2750  gram  of  metallic  nickel  and  cobalt  and 
0.2103  gram  of  sulphates  of  potassium  and  cobalt 
(3X3504,200504)     What  is  the  percentage  of  nickel? 

Ans.     24.52%. 

17.  One  gram  of  a  rock  gave  on  analysis,  combined 
sodium  and  potassium  sulphates  o.  1 50  gram  and  plat- 
inum from  KoPtCle  o.  1 1 2  7  gram.  What  are  the  per- 
centages of  KgO  and  NaoO  ? 

Ans.     K2O  5.46%.     NaoO  2.14%. 

18.  In  an  iron  ore,  ferric  oxide,  alumina  and  phos- 
phoric   acid    were   precipitated    together,    ignited    and 


EXAMPLES.  45 

weighed  as  FcoOg,  AI0O3  and  FeP04.  The  weight 
from  three  grams  of  ore  was  2.4750  grams:  the  iron 
was  found  to  be  50.5%  and  the  phosphorus  0.25%. 
What  is  the  percentage  of  AI2O3?  Ans.     9.74%. 

19.  Which  is  more  economical  for  neutralizing  an 
alkali  60%  HNO3  at6  cents  per  pound,  or  20%  HCl 
at  3  cents  per  pound?  Ans.     HCl. 

20.  Which  is  the  more  economical  oxidizing  agent 
KNO3  at  5  cents  a  pound  or  NaNOg  at  5^  cents  a 
pound?  Ans.     NaNOa. 

2 1 .  Which  is  the  more  economical  neutralizing  agent, 
98%  sulphuric  acid  at  2.85  cents  per  pound,  or  84% 
sulphuric  acid  at  2.35  cents. 

Ans.     84%  at  2.35  cents. 

22.  How  many  pounds  more  of  10%  hydrochloric 
acid  will  it  take  to  dissolve  100  lbs.  of  calcium  carbonate 
than  100  lbs.  of  barium  carbonate?  Ans.     359.05  lbs. 

23.  A  sample  of  coal  was  partially  dried  losing  y% 
of  water.  One  gram  of  this  sample  gave  100  milligrams 
of  barium  sulphate.  What  was  the  percentage  of  sulphur 
in  the  original  sample  ?  Ans.      1.277%. 

24.  Which  is  the  cheaper  source  of  cyanogen,  potas- 
sium cyanide  98%  at  28  cents  a  pound,  or  a  mixture  of 
KCN  68%  and  NaCN  28%  at  30  cents  a  pound  ? 

Ans.     Practically  the  same.     Mixture  preferable. 

25.  An  ore  contained  10%  of  sulphur.  If  in  this  de- 
termination one  tenth  of  the  barium  sulphate  were  reduced 
to  sulphide  by  the  filter  paper,  and  not  reconverted  to 
sulphate,  what  error,  expressed  in  percentage,  would 
result?  Ans.     0.274%. 


CHAPTER  V. 

CALCULATION  AND  USE  OF  FACTORS. 

In  gravimetric  analysis  factor  means  the  number  by 
which  the  weight  of  any  precipitate  or  residue  is  multi- 
pHed  to  give  the  weight  of  the  constituent  desired, 
whether  the  constituent  is  actually  present  or  has  pre- 
viously been  combined  in  a  known  proportion. 

To  calculate  the  factor  for  SO3  in  BaS04  we  have 
X  :  I  :  :  Mol.  wt.  SO3  :  Mol.  wt.  BaS04  or 


Mol.  wt.  SOo  8o.o( 


or or  0.3429. 


Mol.  wt.  BaS04      233.46 

The  factor  is  the  weight  of  SO  3  in  one  gram  of 
BaS04  or  one  hundredth  of  the  percent  of  SO 3  in 
BaSO^. 

To  calculate  the  factors  for  S  and  for  BaO,  we  have 

— 7  =  0.1373  and— ^^^=  0.6571. 

233.46  ^^'^  233.46  ^^ 

Factors  are  used  as  follows : — 
First,  To  simplify  calculations. 

Instead  of  making  the  proportion — 

Mol.  wt.  BaS04 :  Atomic  wt.  S  :  :  wt.  BaS04 :  wt.  S, 
we  have  wt.  BaS04  X  factor  (0.1373)  =  ^^-  S- 
So  that  if  we  have  a  table  of  factors,  the  calculation  re- 
quires only  one  multiplication,  and  this  can  be  changed 
into  an  addition  by  the  use  of  logarithms. 

These  tables  will  be  found  at  the  back  of  the  book. 

(46) 


CALCULATION  AND  USE  OF  FACTORS.         47 

Second,  To  avoid  or  to  greatly  reduce  calculation  by 
weighing  out  a  multiple  of  the  factor  for  analysis. 

If  the  weight  taken  in  grams  equals  the  factor  each 
milligram  of  precipitate  or  residue  equals  o.i%  without 
any  calculation.  When  ten  times  the  factor  is  taken 
each  milligram  equals  0.01%.  When  one  hundred 
times  the  factor  is  used  each  milligram  equals  0.001% 
and  so  on. 

This  can  be  proved  as  follows : — 

Suppose  we  are  to  determine  SO3  in  crystallized 
magnesium  sulphate,  MgS04,7HoO.  Weigh  out  for 
analysis  3.4293  grams  (ten  times  the  factor),  dissolve, 
and  then  precipitate  as  BaS04,  in  the  usual  way;  the 
precipitate  weighed  1.626   grams,  then  we  have 

1.626  X   0.3420    .  .  -n  ^  rcr\ 

^JL^  X  100  =  Percentage  of  SO3 

3429 
but  0.3429  goes  into  3.429  ten  times  and  10  into  100  ten 
times,  so  the  result  in  percentage  is  ten  times  the  weight  in 
grams  or  16.26%,  or  the  precipitate  contained  1626  milli- 
grams, each  of  which  is  therefore  equivalent  to  0.01%. 

Again  the  factor  for  S  in  BaS04  is  0.13733.  If  we 
weigh  out  13.733  grams  of  pig  iron  for  analysis,  each 
milligram  of  BaS04  will  be  equivalent  to  0.001%  S. 
For  suppose  the  BaS04  obtained,  weighed  0.0270  gram  ; 
we  would  have 

0.027x0.13733  X  ^oo  ^  Percent,  of  S  or  0.027%. 

^3-733 
So  27  milligrams  =  27  thousandths  percent  or  each 
milligram  =  0.00 1  percent.  But  this  is  too  large  an 
amount  to  take  unless  the  sulphur  is  extremely  low,  so 
in  practice  some  portion  of  it  is  used  as  ^,  K  or  o-4 ; 
then  the  weight  of  BaS04,  expressed  in  grams,  times 
2,  3,  or  2.5   is  the  percentage   of  sulphur  without  any 


48 


CALCULATION  AND  USE  OF  FACTORS. 


further  calculation.  For  example,  if  for  the  determin- 
ation of  sulphur,  5.4932  grams  is  taken,  the  weight  of 
BaS04  in  grams  divided  by  four  and  multiplied  by  ten 
gives  the  percentage  of  sulphur,  for  the  weight  5.4932 
is  forty  times  the  factor  for  S  in  BaS04.  Suppose  the 
actual  weight  of  BaS04  to  be  0.200  gram,  then 


0.200  X  0.13733 


=  0.005  wt.  S  in  one  gram, 


54932 
and  this  times  100  gives  percentage  or  0.5%  ; 

which  result  is X  10  =  0.5%. 

4 
Inasmuch  as  these  relations  are   confusing  at  first, 
especially  when  the  weights  of  precipitates  are  sometimes 
given  in  grams,  and  sometimes  in  milligrams,  the  follow- 
ing table  may  prove  useful. 


Weight  taken 
in. Grams. 

Weight  Precipitate 
in  Grams. 

Weight  Precipitate  in  Milligrams. 

Factor 

Times  100  =  Percent 

Divided  by  10  =  Percent 
Each  Mg.  =  0.1% 

Factor  X  5. 

Times  20  =  Percent 

Divided  by  50  =  Percent 
EachMg.  =  0.02% 

Factor  X 10 

Times  10  =  Percent 

Divided  by  100  =  Percent 
Each  Mg.  =0.01% 

Factor  X  50 

Times  2  =  Percent 

Divided  by  500  =  Percent 
EachMg  =  0.002% 

Factor  X 100 

Percent  Direct 

Divided  by  1000  =  Percent 
Each  Mg  =  0.001% 

The  use  of  factors  to  save  time  should  not  be  pushed 
too  far,  as  the  time  taken  in  obtaining  an  exact  quantity  of 
some  crystallized  salts  or  of  drillings  may  be  greater  than 


CALCULATION  AND  USE  OF  FACTORS.  49 

that  saved  in  calculation.  The  examples  given  have 
that  objection,  which  however  would  not  apply  to  an  iron 
ore  or  coal  in  which  sulphur  was  to  be  determined,  as 
the  sample  would  be  in  a  finely  pulverized  condition.  It 
takes  but  little  longer  to  weigh  out,  say  exactly  1*37^3 
grams  than  exactly  one  gram,  for  the  time  is  consumed 
in  getting  the  exact  weight,  not  in  putting  the  weights 
on  the  pan :  and  when  an  analysis  is  repeatedly  made, 
even  this  time  can  be  saved  by  using  a  special  weight, 
the  most  convenient  m.ultiple  of  the  factor. 

This  system  can  be  very  easily  applied  also  to  volu- 
metric work,  here  the  amount  to  be  weighed  out  will  be 
a  multiple  of  the  standard  (titer)  of  the  solution  used. 
Suppose  we  have  a  solution  of  KMn04  i  c.c.  =  0.0056 
gram  of  iron.  If  we  take  for  analysis  0.56  gram  of 
iron  ore  and  55.1  c.c.  of  KMn04  are  used,  the  ore  con- 
tains 55.1%  of  iron  for 

SS-I    X   0.0056  ./  ^^     ny 

^^ ^  X  100=  55-i%- 

0.56 

So  that  when  we  take  for  the  test  a  weight  equal  to 
hundred  times  the  standard  of  the  solution,  each  cubic 
centimeter  equals  one  percent  and  the  burette  reading 
gives  percentage  direct. 

One  hundred  times  the  standard  may  not  be  the  most 
convenient  quantity  to  take,  but  this  can  be  readily  be 
adjusted  to  the  particular  titration,  bearing  in  mind  that, 
when 

50  times  the  standard  is  taken  i  c.c.  =  2%. 
100      *'       ''         ''  *'  I  c.c.  =  1%. 

200      "        "         ''  **  ICC.  =  0.5%. 

and  so  on. 

The  plan  of  adjusting  the  weight  taken  to  the 
strength  of  the  solution  presents  many  advantages  ove- 


50  CALCULATION  AND  USE  OF  FACTORS. 

the  system  of  adjusting  the  solution  to  a  certain  weight 
for  analysis,  when  a  portion  is  used  for  a  single  deter- 
mination. This  is  particularly  the  case  when  the  standard 
solution  changes  in  strength  and  has  to  be  frequently  re- 
standardized.  If  we  always  use  one  burette,  which  is 
uniformly  calibrated,  for  the  particular  solution  and 
weigh  out  a  multiple  of  the  standard  for  analysis,  that 
burette  will  read  percentage :  and  we  do  not  have  to 
make  up  the  solution  to  a  certain  arbitrary  strength, 
which  necessitates  accurate  measuring  apparatus,  flasks, 
etc.,  which  agree  with  the  burette.  Then  if  on  standing 
the  solution  evaporates  or  undergoes  a  partial  decom- 
position, it  must  be  readjusted  to  read  percentage  when 
an  arbitrary  weight  is  taken,  while  by  weighing  out  a 
multiple  of  the  new  standard  much  trouble  is  saved  with 
no  sacrifice  of  accuracy.  It  is  not  to  be  inferred  how- 
ever that  this  system  is  advantageous  in  all  cases,  but 
only  for  separate  determinations  such  as  iron,  copper, 
zinc,  etc. 

The  most  important  application  of  the  use  of  factors, 
is  the  assay  ton  system  devised  by  Prof.  Chandler.  In 
assaying  ores  of  gold  and  silver,  the  precious  metals  are 
always  reported  in  ounces  Troy  per  ton  of  2000  pounds 
avoirdupois.  If  a  suitable  number  of  grams,  twenty-five 
for  instance,  is  taken  and  the  gold  and  the  silver  weighed 
in  milligrams,  a  long  calculation  is  required  to  convert 
the  results  to  ounces  per  ton.  If  however  an  assay  ton 
29.1666  grams,  is  used,  the  result  in  gold  or  silver  as 
weighed  in  milligrams  Is  ounces  to  the  ton  without  any 
calculation. 

The  assay  ton  is  derived  as  follows : — 
One  pound  avoirdupois  contains  7000  grains. 
One  ton,  2000  lbs.,  contains  14,000,000  grains. 
One  ounce  Troy  contains  480  grains. 


CALCULATION  AND  USE  OF  FACTORS.         51 

Therefore  14,000,000  -^  480  or  29,166.6  is  the  number 
of  Troy  ounces  in  2000  lbs.  avoirdupois  ;  so  if  we  take 
this  number  of  milHgrams  for  assay  (29,166.6  +),  each 
milHgram  of  gold  or  of  silver  found  is  an  ounce  Troy  to 
the  ton  in  the  ore  :  for 

I  :  29,166.6  +  :  :  480  :  14,000,000. 

Therefore  the  assay  ton  is  29,166.6  +  milligrams  or 
29.1666  +  grams. 


EXAMPLES. 


1.  Calculate  the  factors  for  NagO  in  Na2S04  and  in 
NaCl.  Ans.     0.4368  and  0.5308. 

2.  Calculate  the  factors  for  Zn  in  ZnO,  in  ZnNH4 
PO4  and  in  Zn^^^Orj, 

Ans.     0.80345,  0.3664  and  0.4291. 

3.  Calculate  the  factors  for  As  in  As 2 S3,  in  MggAsgOy 
and  Ag3As04.  Ans.     0.6093,  0.4827  and  0.1621. 

4.  Calculate  the  factors  for  N,  NH4  and  NH4CI,  for 
weight  of  platinum  from  ignition  of  (NH4)2PtCl6. 

Ans.     0.1441,  0.1752  and  0.5495. 

5.  What  are  the  factors  for  conversion  of  Cr  into 
CrgOe,  into  PbCr04  and  into  BaCr04  ? 

Ans.     1.4606,  6.1996  and  4.8656. 

6.  (a)  What  is  the  factor  to  convert  weight  of  BaS04 
to  corresponding  weight  of  H2SO4  ? 

(b)  What   is   the  factor   to   convert   percentage   of 
K2O  to  percentage  of  K2SO4  ? 

Ans.     (a)  0.42010.     (b)   1.8490. 

7.  What  are  the  factors  for  KgO  and  for  KCl  in 
platinum  from  KgPtCle  ? 

Ans.     0.48409  and  0.76591. 

8.  What  are  the  factors  for  the  iron  in  Fe2  03  and  in 
Fe304?  Afis.     0.6996  and  0.7238. 

9.  One   gram  of  an  antimony   ore   was   taken    for 
analysis  and  the  Sb2  04  resulting  weighed  0.3478  gram  : 

(52) 


EXAMPLES.  53 

by  mistake   the  factor  for  SboSg  was  used.     What  was 
the  error  in  percentage  of  antimony  ? 

Ans.     2.629%. 

10.  An  alloy  contains  10%  of  antimony.  One  gram 
of  this  alloy  gives  0.2895  gram  of  combined  oxides  of 
antimony  and  tin  (Sbo04  and  SnOo).  What  would  be 
the  error  in  percentage  of  tin  if  only  the  factor  for  Sb204 
were  used  in  the  calculation  ?  Ans.     0.028%. 

1 1.  What  weights  of  spiegel  must  be  taken  for  ana- 
lysis so  that  each  milligram  of  MngPoO;  shall  equal  re- 
spectively 0.2%,  0.1%  and  0.05%? 

Ans.     0.19365,  0.3873  and  0.7746. 

12.  What  weights  of  sphalerite  must  be  used  so  that 
each  milligram  shall  equal  0.1%  when  the  zinc  is 
weighed  as  ZnO,  as  ZnNH4P04  and  as  ZuoP^Oy  ? 

Ans.     0.8035,  0.3664  and  0.4291. 

13.  What  weight  of  steel  must  be  taken  so  that,  when 
the  phosphorus  is  weighed  as  MgoPoO:,  the  weight  in 
grams  shall  be  percentage  direct?       Ans.     27.8375. 

14.  What  weight  of  pig  iron  must  be  taken  so  that  the 
weight  of  BaS04  in  grams  shall  be  percentage  of  sul- 
phur when  multiplied  by  three.  Ans.     ^.^yjG. 

15.  What  weights  of  lead  ore  must  be  used  so  that 
each  milligram  of  PbS04  shall  equal  0.05%,  0.1%  and 
0.2%?  Ans.      1.3658,  0.6829  and  0.3415. 

16.  What  weights  of  limestone  must  be  taken  so  that 
each  milligram  of  CaO  equals  0.1%  of  CaO  ?  So  that 
each  milligram  of  CaS04  equals  o.  1%  of  CaO  ? 

Ans.     I  and  0.4120. 

17.  Given  a  solution  of  KoCroOy,  i  c.c.  =  0.0054 
gram  of  iron.  How  much  ore  must  be  taken  so  that 
each  c.c.  shall  read  0.1%,  0.2%,  0.3%,  0.5%,  0.7% 
and  i.%?  Ans.     5.4,  2.7,  1.8,  1.08,  0.7714,  0.54. 


54  EXAMPLES. 

i8.  Given  a  solution  of  K4Fe(CN)tj,  i  c.c.  =  0.00896 
gram  of  zinc.  How  much  ore  must  be  taken  so  that 
each  c.c.  shall  equal  2%,  1%,  0.5%,  o.i%? 

0.4480,  0.8960,  1.792,  8.960. 

19.  Calculate  the  ounces  Troy  per  ton  avoirdupois 
in  the  following — 

(a)  One  tenth  assay  ton  of  copper  matte  gave  gold 
1.22  mgs.,  silver  30.94  mgs. 

(b)  Four  assay  tons  of  siliceous  ore  gave  gold  o.  1 2 
mg.,  silver  0.52  mg. 

(c)  One  sixth  assay  ton  of  a  zinc  blende  gave  gold 
0.02  mg.,  silver  37.72  mgs. 

Ans.: 
,  .  ^  12.2  ozs.  Au.  ^1 N  (  0.03  oz.Au.  ^  N  (  o.  12  oz.  Au. 
\  309.4 ozs.  Ag.^  ^  C  0-I3  oz.  Ag.  ^  ^  (  226.32  ozs.  Ag. 

20.  Calculate  the  number  of  grams  in  an  assay  ton 
for  a  long  ton  (2240  pounds). 

Ans.      32.6666  grams. 


CHAPTER  VI. 

CALCULATIONS   OF   VOLUMETRIC   ANALYSIS. 
PART  I.     GENERAL. 

NORMAL  SOLUTIONS. 

A  normal  solution  is  one  which  contains  the  hydro- 
gen equivalent  of  the  active  constituent  in  grams  per 
liter.  That  is,  the  amount  in  a  liter  which  brings  into  re- 
action 1.008  grams  of  hydrogen,  8  grams  of  oxygen  or 
their  equivalents,  whether  the  reaction  be  one  of  oxida- 
tion, reduction,  precipitation  or  saturation. 

This  definition  is  used  by  Mohr  and  by  Sutton  and  is 
also  in  very  general  use  in  America.  In  the  examples 
given  in  this  book  this  system  is  used.  There  are  how- 
ever other  definitions  of  normal  solutions,  which  have 
caused  considerable  confusion:     They  are : 

First.  The  molecular  weight  in  grams  per  liter.  This 
definition  possesses  some  advantages,  for  example  ac- 
cording to  Mohr  s  definition  we  can  have  two  normal 
solutions  of  KMn04  of  different  strengths  depending  on 
whether  it  is  used  in  a  neutral  or  in  an  acid  solution ; 
while  according  to  the  "  molecular  weight  in  grams  "  this 
can  not  happen,  on  the  other  hand  Mohr's  definition  has 
the  great  advantage  that  all  normal  solutions  are  equiva- 
lent cubic  centimeter  for  cubic  centimeter.  Solutions 
containing  the  molecular  weight  in  grams  per  liter  are 
now  much  used  in  physical  chemistry  under  the  name 
gram-molecule. 

Second.  In  the  Gay-Lussac  method  for  the  determ- 
ination of  silver,  used  in  mints  and  assay  offices,  normal 

55 


56  NORMAL  SOLUTIONS. 

salt  solution  Is  the  name  given  to  a  solution  of  such 
strength  that  one  hundred  cubic  centimeters  will  exactly- 
precipitate  one  gram  of  silver.  This  is  a  special  use  of 
the  word  normal  and  should  not  be  connected  in  any 
way  with  the  equivalent  system  of  volumetric  analysis. 
Under  the  system  we  have  adopted,  it  is  absolutely 
essential  to  have  a  correct  understanding  of  '*the  hydro- 
gen equivalent  in  grams."  Suppose  we  wish  to  make  a 
normal  solution  of  sodium  hydroxide ;  we  have  the  re- 
action— 

NaOH+HCl=NaCl  +  HoO. 

What  is  the  amount  necessary  to  bring  into  reaction 
one  equivalent  of  hydrogen  ?  Evidently  the  molecular 
weight  in  grams  of  NaOH  (40.058)  for  this  is  the  amount 
necessary  to  react  with  36.458  grams  of  HCl  and  so 
bring  into  reaction  1.008  grams  of  hydrogen. 

If  we  wish  a  normal  solution  of  NaoCOs  we  have 
NaoC03+  2HCI  =  2NaCl+  H0O  +  COJ  It  is  evident 
that  one  moleculeofNaoCOg  brings  into  reaction  two 
atoms  of  hydrogen,  hence  a  normal  solution  of  NaoCOg 
will  be  one  half  of  the  molecular  weight  or  53.05  grams 
per  liter. 

Similarly  a  normal  solution  of  hydrochloric  acid  will 
contain  the  molecular  weight,  36.458  grams  per  liter,  of 
sulphuric  acid  one  half  the  molecular  weight,  49.038 
grams,  and  of  orthophosphoric  acid  one  third  the  molec- 
ular weight  in  grams  or  32.6747  grams. 

One  cubic  centimeter  of  any  of  these  acid  solutions 
will  exactly  saturate  one  cubic  centimeter  of  any  of  the 
alkaline  solutions. 

In  reactions  of  precipitation  the  same  rule  applies 
we  have, — 

AgNOa  +  HCl  =  AgCl  +  HNO3  or 
AgNOa  +NaCl  =  AgCl  +  NaNOg 


NORMAL  SOLUTIONS.  57 

A  normal  solution  of  silver  nitrate  contains  the  molecu- 
lar weight  in  grams,  169.97,  per  liter:  for  according  to 
the  first  reaction  1.008  grams  of  hydrogen  would  be 
actually  brought  into  reaction  by  a  liter  of  this  solution 
and  in  the  second,  the  weight  of  sodium  equivalent  to 
1 .008  grams  of  hydrogen.  Similarly  with,  NH4SCN  + 
AgN03=AgSCN+NH4N03,  a  normal  solution  will 
contain  the  molecular  weight  of  ammonium  thiocyanate 
in  grams  per  liter  or  76.17  grams.  To  precipitate  a  salt 
of  a  diatomic  element  such  as  BaCr04,  we  have  a  normal 
solution  equal  to  one  half  the  molecular  weight  in  grams 
per  liter;  for  KgCrO^,  97.2  grams;  for  Na2Cr04.  81. i. 
grams. 

For  convenience  we  designate  a  normal  solution  by 
N,  tenth  normal  by  %,  one  hundreth  normal  by  ^/joo 
and  so  on. 

In  the  original  definition  of  normal  solution  appears 
"equivalent  to  one  gram  of  hydrogen."  This  can  be 
retained  with  perfect  accuracy  if  the  atomic  weights  used 
are  calculated  on  the  basis  of  hydrogen  as  unity  ;  but 
as  the  ratio  of  hydrogen  to  oxygen  has  been  found  to 
be  I  :  15.88  and  not  1:16  we  must  adopt  either  H  =  i 
or  O  =  16  as  a  basis.  O  =  16  is  most  generally  used  at 
present  so  to  make  our  definition  of  normal  comparable 
with  the  present  atomic  weights,  instead  of  equivalent 
to  one  gram  of  hydrogen,  we  must  substitute,  equivalent 
to  1.008  grams  of  hydrogen  or  better  equivalent  to  8 
grams  of  oxygen.  The  advantage  of  the  latter  definition 
will  appear  in  the  following  explanations  of  normal  solu- 
tions for  oxidation  and  reduction. 

When  ferrous  iron  is  oxidized  by  potassium  dichro- 
mate  we  have  the  reaction  KsCrgOy  +  6FeCl2  +  14HCI 
=  6FeCl3  +  2KCI  +  2CrCl3  +  7H2O:  the  essential  oxi- 
dation in  this  reaction  can  be  expressed  as  follows  — 


58  NORMAL   SOLUTIONS. 

KgCrgOy  can  be  regarded  as  breaking  up  into  KgO. 
CrgOg  and  3O  and  6FeO  +  3O  =  sFeaOg.  Three 
atoms  of  oxygen  from  the  KgCrgOy  oxidize  six  atoms  of 
ferrous  iron  as  chloride  to  the  ferric  condition,  or  in  other 
words  an  iron  requires  one  half  an  oxygen,  equivalent  to 
one  hydrogen :  so  for  a  normal  solution,  the  molecular 
weight  of  KgCrgOy  in  grams  must  be  divided  by  six, 
since  each  molecule  oxidizes  six  atoms  of  iron  each  equi- 
valent to  one  atom  of  hydrogen  or  one  half  an  atom  of 
oxygen  or  expressed  in  the  form  of  a  proportion  we  have  : 

Mol.  wt.  KgCrgOy  :  X  :  :  3  X  16  :  8.    x  =  49.o83  grams. 

With  KMn04  we  have  2KMn04  +  ioFeS04  + 
8H2S04  =  5Fe2  (804)3  +K2SO4  +2MnS04  +8H2O. 

The  essential  oxidation  is  MnaOy-]-  ioFeO  =  5Fe2  03 
-f-  2MnO  or  2KMn04  give  5  oxygen.  To  calculate  the 
oxygen  equivalent  we  have,  2  (Mol.  Wt.  KMn04)  =316.3 
grams,  gives  oxygen,  5  X  16  =  80  grams,  hence  316.3  : 
x  : :  80  :  8  or  a  normal  solution  of  KMn04  contains  31.63 
grams  per  liter  when  used  in  an  acid  solution. 

The  following  may  make  the  calculation  of  N  KMn04 
more  easily  understood.  2KMn04  give  KgO,  2MnO 
and  5O  :  the  essential  part  of  the  oxidation  of  iron  is 
2FeO  +  0  =  Fe203. 

Two  molecules  of  permanganate  furnish  5  atoms  of 
oxygen  :  two  molecules  of  ferrous  oxide,  each  containing 
one  iron,  require  one  oxygen,  equivalent  to  two  hydro- 
gen, so  that  twice  the  molecular  weight  in  grams  of 
permanganate  furnish  the  equivalent  of  80  grams  of 
oxygen  or  10  -(-  grams  of  hydrogen.  So  the  normal  so- 
lution becomes 

2  X  158.15 
10 
or  31.63  grams  per  litre. 


NORMAL  SOLUTIONS.  59 

In  the  Volhard  method  for  manganese  we  have  in  a 
neutral  solution  the  following  reaction 

2KMn04  +  3MnS04  +  2ZnO  = 

SMnOs  +  K2SO4  +  2ZnS04 

the  essential  oxidation  is  MngOy  +  3MnO  =  sMnOg  or 
two  molecules  of  KMn04  give  up  under  these  conditions 
three  atoms  of  oxygen,  so  to  get  a  normal  solution  we 
have  316.3  :  X  : :  48  :  8  or 

X  =  2  Mol.  Wt.  -^  6  =  52.717  grams  per  liter. 

This  is  one  of  the  very  few  cases  where  the  molecu- 
lar weight  definition  of  normal  seems  preferable,  as  here 
we  have  two  normal  solutions  of  different  strengths  de- 
pending on  the  reaction.  As  permanganate  is  usually 
used  in  an  acid  solution,  unless  expressly  stated,  normal 
means  31.63  grams  per  liter. 

Normal  reducing  solutions  are  not  much  employed, 
but  the  following  will  serve  as  an  illustration.  We  have 
the  reaction 

SnClg  +  sFeClg  =  SnCl4  +  2FeCl2  : 

one  molecule  of  SnClg  removes  two  atoms  of  chlorine  or 
what  is  the  same  thing  reduces  two  atoms  of  iron  to  the 
ferrous  condition.  Now  two  chlorine  correspond  to  one 
oxygen  ;  hence  one  half  the  molecular  weight  in  grams 
of  stannous  chloride  will  give  a  normal  solution  or 

189.9  :  X  : :  16  :  8.     x  =  94.95  grams. 

Or  more  simply  in  this  case,  one  molecule  of  SnClg  is 
equivalent  in  reducing  power  to  two  atoms  of  hydrogen, 
so  the  normal  solution  is  one  half  the  molecular  weight 
in  grams. 


60  NORMAL   SOLUTIONS. 

To  ascertain  the  strength  of  any  normal  solution  the 
reaction  in  which  it  is  to  be  used  must  always  be  consid- 
ered and  in  this  way  all  confusion  will  be  avoided. 

For  example,  potassium  dichromate  can  be  used  for 
precipitating  barium  or  lead,  as  well  as  for  oxidation  as 
already  described. 
KgCraO^  +  2BaCl2  +  2NH4OH 

=  2BaCr04  +  2KCI  +  2NH4CI  +  HsO^ 
Here  one  molecule  brings  into  reaction  two  atoms  of 
barium,  each  equivalent  to  two  hydrogen,  so  the  normal 
potassium  dichromate  for  precipitation  is  one  quarter  the 
molecular  weight  in  grams  per  liter  or  73.625  grams  ;  not 
49.083  grams,  the  normal  solution  for  oxidation. 

An  advantage  of  the  *'  equivalent "  definition  of  nor- 
mal, which  deserves  attention,  is  that  when  the  formula 
of  a  compound  is  altered  the  equivalent  is  not ;  so  that 
whether  you  consider  potassium  permanganate  as  K2Mn2 
Og  or  KMn04,  arsenious  oxide  as  AsgOg  or  AS4O6  or 
ferric  chloride  as  FegClg  or  FeClg,  the  number  of  grams 
per  liter  of  a  normal  solution  will  remain  the  same.  This 
is  made  clear  by  the  following  reactions. 
K2Mn208  +  5Fe2(S04)2  +  8H2SO4  = 

5Fe2(S04)3  +  K2SO4  +  2MnS04  +  8H2O 
Now  written 
2KMn04  +  ioFeS04  +  8H2SO4  = 

5Fe2(S04)3  +  K2SO4  +  2MnS04  -f  8H2O 

In  either  case  the  same  weight  of  potassium  perman- 
ganate oxidizes  the  same  weight  of  iron  and  normal  solu- 
tions will  be  identical  for  N  K2Mn2  08  will  be  one  tenth 
the  molecular  weight,  31.63  grams,  and  N  KMNO4  one 
fifth,  3 1 .63  grams.     The  same  applies  to  ferrous  sulphate. 

In  using  normal  solutions  the  following  rules  are  im- 
portant.    Weigh  out  one  tenth  the  equivalent  in  grams 


NORMAL    SOLUTIONS.  61 

of  the  substance  to  be  determined  and  the  burette  reading 
gives  the  percentage  directly.  With  ^/jo  solutions  weigh 
out  one  hundreth  of  the  equivalent  in  grams  and  the 
burette  shows  the  percentage  directly.  To  make  this 
clear,  suppose  we  have  a  normal  sulphuric  acid  solution 
and  we  wish  to  titrate  a  sample  of  caustic  potash  and 
to  express  the  results  in  percentage  of  KoO.     N  H0SO4 

contains      '        grams  per  liter  or  i.  c.c.  =  0.049038  gram 

ofH2S04.     The  equivalent  K2O  is  ^^  or  47.15  ;  so  N 

H2SO4  is  equivalent  to  0.04715  gram  K2O  per  c.c. 
One  tenth  the  equivalent  in  grams,  4.715,  is  weighed  out, 
dissolved  and  titrated,  requiring  78.2  c.c.  of  N  H2SO4. 
Obtain  the  percentage  by  going  through  the  ordinary 
calculation 

It  is  seen  that  this  is  only  another  way  of  expressing  the 
fact  that  when  one  hundred  times  the  standard  is  taken 
for  analysis,  the  burette  gives  percentage  direct/*^ 

Besides  normal  solutions  we  have  standard  solutions, 
that  is  solutions  of  known  strength.  Normal  solutions 
are  always  standard  but  standard  solutions  are  not 
always  normal.  1  he  salt  solution  used  in  the  mints  is 
standard,  but  only  normal  in  a  particular  sense  of  the 
word.  Most  of  the  solutions  used  in  volumetric  analy- 
sis are  standard  and  only  approximately  normal,  tenth 
normal  or  one  hundreth  normal.  The  question  of  the 
strength  of  solutions  and  whether  systematic  or  not  is 
largely  a  matter  of  individual  preference,  limited  by  the 

(*)  A  table  of  the  more  common  N  and  N/10*solutions  will  be  found  at  the 
back  of  the  book. 


62  ADSUSTMENT  OF  SOLUTIONS. 

solubility  of  the  reagent  and  the  requirements  of  the  par- 
ticular analysis  to  be  performed. 

In  the  calculations  of  volumetric  as  in  gravimetric 
analysis  the  basis  of  calculation  is  an  equation.  Unfort- 
unately there  are  some  cases  in  which  the  reaction  can 
not  be  expressed  by  a  simple  equation  and  in  these  cases 
standard  not  normal  solutions  must  be  used. 

ADJUSTMENT    OF    SOLUTIONS. 

As  this  is  in  no  sense  a  book  on  volumetric  analysis 
a  discussion  of  the  merits  of  the  different  liters  in  practi- 
cal use  is  out  of  place  and  the  calculations  are  given  so 
as  to  apply  to  whatever  unit  is  taken  for  the  volumetric 
apparatus. 

Suppose  we  have  made  up  ten  liters  of  sulphuric 
acid  approximately  normal  and  find  on  standardizing  that 
the  strength  is  i  c.c.  =  0.050  gram  H2SO4  and  that  in 
making  the  test  we  have  used  50  c.c.  of  the  solution. 
We  have  left  9,950  cubic  centimeters  and  the  solution  is 
too  strong.  To  find  the  amount  of  water  to  add  to  make 
it  normal :  let  x  be  the  volume  required  so  that  the 
solution  shall  be  normal,  then  x  times  0.049038  is  the 
weight  of  H0SO4  contained  therein  ;  9,950  X  0.050  is  the 
weight  of  H2SO4  in  our  solution.  These  quantities  are 
equal  as  they  will  not  be  affected  by  dilution  therefore 
we  have 

9^950  X  0.050  =  X  X  0.049038. 
X  =  10,145.2  c.c.     So  the  amount  of  water  to  be  added 
is    10,145.2  —  9,950  or    195.2  c.c.     (These  cubic  centi- 
meters must  accord  with  the  liter  adopted.) 

Suppose  the  solution  gave  on  standardizing  i  c.c.  = 
0.048  gram  H2SO4.  The  solution  is  too  weak.  By  a 
similar  method  we  can  find  the  amount  of  water  in  excess 
and  then  add  the  proper  quantity  of  sulphuric  acid,  we  have 


ADJUSTMENT   OF   SOLUTIONS.  63 

9950  X  0.048  =  X  X  0.049038.     X  =  9739.4. 

The  excess  of  water  is  therefore  9950  —  9739.4  or  210.6 
c.c.  For  the  amount  of  sulphuric  acid  to  be  added  see 
Chapter  VII. 

It  will  be  readily  seen  that  it  is  much  more  convenient 
to  make  up  the  solutions  too  strong  rather  than  too 
weak. 

If  we  have  a  solution  of  say  NasCOg  which  is  too  weak  ; 
find  out  the  amount  of  water  in  excess,  then  calculate  the 
amount  of  the  salt  for  this  volume  of  water  and  add  it. 
The  change  in  volume  due  to  the  solution  of  the  salt  can 
be  neglected. 

Example.  A  solution  of  NasCOg  was  found  to  have 
a  strength  i  c.c.  =  0.047  g^^^  of  H2SO4.  How  much 
NaoCOa  must  be  added  to  two  liters  to  make  it  normal? 

2000  X  0.047  =  X  X  0.049038.     x  =  1,916.9. 

There  are  83.1  c.c.  of  water  in  excess.  Now,  a  normal 
solution  of  NagCOg  contains  }4  the  molecular  weight  in 
grams  per  Hter  or  53.05  grams,  so  the  amount  to  be 
added  is 

53.05  :  y  : :  1000  :  83.1.     y  =  44085  grams. 

We  may  have  two  standard  solutions  of  a  reagent 
and  desire  to  make  a  third  of  a  certain  intermediate 
strength  —  for  example  suppose  we  have  two  solutions  of 
NH4SCN 

(a)  I  c.c.  =  0.01275  gram  of  silver 

(b)  I  c.c.  =  0.00447  gram  of  silver 

and  we  wish  to  make  one  liter  of  a  solution  equivalent 
to  10  mgs.  of  silver  per  c.c.  How  much  of  each  shall 
we  mix.  Let  x  be  the  number  of  c.c.  of  (a)  ;  then  1000  — 
X  =  number  of  c.c.  of  (b)  :  and  we  have 


64  EXCESS   NECESSARY   TO   AFFECT  THE   INDICATOR. 

X  (0.01275)  +  (1000  — x)  0.00447  =  10 
(as  one  liter  will  contain    10  grams  if   i   c.c.  contains 
10  mgs.)     This  gives  667.9  c.c.  of  (a)  and  332.1  c.c.  of 

(b). 

If  we  have  a  solution  of  KMn04  containing  2  grams 
per  liter  and  a  ^/s  solution  of  the  same,  to  make  two 
liters  of  a  ^/lo  KMn04.  Let  x  be  the  volume  of  the  2 
grams  solution  in  liters  and  (2  —  x)  the  volume  of  the  ^/g 
solution  ;  then  2  x  +  (2  -x)  6.326  =  2  (3.163)  and  x  = 
1.4623  liters,  (2  —x)  =  0.5377  liter. 

SUBTRACTION    OF    EXCESS    NECESSARY  TO    AFFECT  THE 

INDICATOR. 

In  some  titrations,  conspicuously  that  of  zinc  by  potas- 
sium ferrocyanide  and  that  of  lead  by  ammonium  molyb- 
date,  the  end  point  is  not  very  sharp  and  a  considerable 
excess  of  the  standard  solution  is  used  before  the  indica- 
tor is  affected.  This  amount  should  be  determined  by 
running  a  blank  test  under  the  same  conditions  as  to 
temperature,  bulk,  acidity  &c.,  and  should  be  subtracted 
from  the  burette  reading  before  calculating  the  results  ; 
for  this  amount  is  the  excess  necessary  to  affect  the  indi- 
cator and  is  required  when  there  is  no  metal  present. 
Other  conditions  remaining  the  same  this  excess  is  usu- 
ally proportional  to  the  bulk,  but  has  to  be  determined 
for  each  set  of  conditions. 

This  subtraction  of  the  excess  necessary  to  turn  the 
indicator  can  be  omitted  when  the  solution  titrated  cpn- 
tains  nearly  the  same  weight  of  metal,  as  well  as  the 
same  volume  &c.,  as  was  used  in  standardizing,  for  then 
this  appears  in  the  lower  standard  of  the  solution  and  no 
appreciable  error  is  introduced.  These  conditions  are 
however  almost  impossible  to  attain  with  a  variety  of 
ores,   so  that  the  best  plan   is  to  make  the  subtraction 


EXCESS   NECESSARY  TO  AFFECT  THE   INDICATOR.  65 

from   every  reading,    for  a   color  and  set  of  conditions 
found  satisfactory. 

The  following  will  illustrate  this  fact — 

A  solution  of  crystallized  potassium  ferrocyanide  was 
made  up  for  the  titration  of  zinc  ores,  approximately,  44 
grams  per  liter.     To  standardize;  200  mgs.  of  pure  zinc 
were  titrated  under  the  standard  conditions,  using  uran- 
ium acetate   as   an   indicator.     Then  a  blank  was  run 
under  the  same  conditions. 
0.200  gram  ofzinc+HgO,  HCl  &c  required  20.45  c.c. 
Blank  (HoO,  HCl  &c)  required  0.45  c.c. 
0.200  gram  of  zinc  required  20  c.c. 
ICC.  =  0.0 10  gram  of  zinc. 

Disregarding  the  blank  the  strength  becomes 
I. c.c.  =  0.00978  gram  of  zinc. 

We  now  titrate  one  gram  of  an  ore  and  use  21  c.c.  : 
using  the  blank  this  gives  20.55%:  disregarding  the 
blank  and  using  the  lower  standard  gives  20.538%. 
Results  close  enough  for  ordinary  work.  But  suppose 
we  now  titrate  a  rich  ore,  say  one  gram  requires  un- 
der the  same  conditions  56  c.c.  :  using  the  blank  we 
g^t  55.55%  and,  disregarding  the  blank  and  using  the 
lower  standard,  we  get  54.768%.  This  involves  an 
error  not  to  be  disregarded  under  any  circumstances. 

The  excess  necessary  to  turn  the  indicator  is  worthy 
of  consideration  in  all  titrations.  It  may  be  too  small  to 
be  subtracted  in  some  cases,  but  there  are  many  in  which 
this  error  is  improperly  neglected.  Some  of  the  discrep- 
ancies in  alkalimetric  titrations  by  the  use  of  different 
indicators,  might  be  reduced  if  not  eliminated  by  making 
allowance  for  the  excess  necessary  to  give  a  certain  color 
with  the  indicator. 


EXAMPLES. 


1.  Calculate  the  number  of  grams  per  liter  to  give 
normal  solutions  of  oxalic  acid,  of  tartaric  acid,  of  acetic 
acid  and  of  citric  acid. 

Ans.  H2C2O4,  45.008  grams;  H2C4H4O6,  75.024 
grams;  HC2H3O2,  60.032  grams;  HgCgHgOy,  64.021 
grams. 

2.  Calculate  the  amounts  necessary  to  make  two 
liters  of  ^/lo  NaCl  and  NaBr. 

Ans,     NaCl  11.7  grams  ;  NaBr  20.602  grams. 

3.  Given  a  solution  of  K2Cr2  07,  i  c.c.  =  0.0042 
gram  Fe,  how  many  grams  per  liter  of  KMn04  will  give 
a  solution  of  equal  strength  ? 

Ans.     2.3764  grams. 

4.  How  many  grams  of  K2Cr207  per  liter  will  give 
a  solution  equivalent  to  a  KMn04  solution  containing 
2.5  grams  per  liter?  Ans.     3.8795  grams. 

5.  Two  liters  of  KMn04  solution  were  made  up 
approximately  ^/g.  It  was  found  on  standardizing  that 
the  strength  was  i  c.c.  =  0.01175  gram  Fe  and  48.2  c.c. 
were  used.  How  much  water  must  be  added  to  what 
remains  to  make  it  ^/g?  Ans.     95.5  c.c. 

6.  How  many  grams  per  liter  of  KMn04  will  give 
a  solution  of  such  strength  that  i  c.c.  =  1%  of  iron  when 
0.50  gram  is  taken  for  analysis? 

Ans.     2.8292  grams. 

(66) 


EXAMPLES.  67 

7.  How  many  grams  per  liter  of  K2Cr207  will  give 
a  solution  of  such  strength  that  i  c.c.  =  1%  of  iron  when 
0.60  gram  is  taken  for  analysis  ? 

Ans,     5.2684  grams. 

8.  How  many  grams  of  NagSgOg,  5H2O  per  liter 
will  give  a  solution  such  that  i  c.c.  =  0.5%  Cu,  when 
one  gram  is  taken  for  analysis  ? 

Ans.     19.520  grams. 

9.  How  much  water  must  be  added  to  two  liters  of 
KMn04,  I  c.c.  =  0.0065  gram  iron,  to  make  it  ^/lo? 

Ans.     325.6  c.c. 

10.  How  much  water  must  be  added  to  three  liters 
of  K4Fe(CN)6,  I  c.c.  =0.0115  gram  zinc,  to  make  it 
read  percentage  directly  when  one  gram  is  taken  for 
analysis  ?  Ans.     450  c.c. 

11.  How  many  grams  of  K4Fe(CN)6,  3H2O  must 
be  added  per  liter  to  a  solution  of  such  strength  that" 
I  c.c.  =  0.0048  gram  of  zinc,  to  make  its  strength  5  mil- 
ligrams of  zinc  per  c.c?  Given  the  composition  of  the 
precipitate  K2Zn3(Fe(CN)6)2.        Ans.     0.8620  gram. 

12.  How  much  water  must  be  added  to  50  grams  of 
HCl  sp.  gr.  1. 1 5  containing  29.57%  H CI  by  weight  ta 
make  it  normal  ? 

N  HCl  =  0.036458  gm.  HCl  per  c.c. 

1. 15  sp.  gr.  HCl  =  29.57%, 

50  X  0.2957  =  14785  grams  HCl, 

weigfht  , 

^ —  =  volume, 

sp.  gr. 

/.  — —  =  4^.48  c.c.  volume  of  HCl. 
1. 15 


6S  EXAMPLES. 

Let  X  =  volume  in  c.c.  of  N  HCl  containing  14.785 
gms.  HCl,  then 

36.458  :  14.785  : :  1000  :  x.     x  =  405.53  c.c, 
405.53  -  43.48  =  362.05  c.c.      Ans. 

13.  How  much  water  must  be  added  to  100  grams 
of  nitric  acid  (1.4  sp.  gr.)  containing  65%  HNO3  by 
weight  to  make  it  ^72  ?  Ans.      1990.4  c.c. 

14.  How  many  grams  of  sulphuric  acid  containing 
^J%  H2SO4  by  weight  must  be  diluted  to  two  liters  to 
give  a  fifth  normal  solution  ?       Ajis.     22.5462  grams. 

15.  How  much  NaOH  must  be  added  to  1.890  liters 
of  NaOH  solution,  whose  strength  is  i  c.c.  =  0.045 
gram  H2SO4,  to  make  it  normal?  How  much  water  to 
make  it  ^/g  ? 

Ans,     6. 172  grams  NaOH.     1 581.3  c.c.  water. 

16.  What  volumes  of  ^/g  and  of  ^/jq  H2SO4  must  be 
mixed  to  give  two  liters  of  ^/s  H2SO4  ? 

Ans.     ^  liter  ^72  and  lyi  liters  ^/iq. 

17.  A  ^/lo  iodine  solution  is  2%  too  strong.  How 
many  cubic  centimeters  of  a  ^1  m  iodine  solution  must  be 
added  to  one  liter  to  make  it  right  ? 

Ans.     22.22  c.c. 

18.  Under  certain  conditions,  in  the  titration  of  lead 
by  ammonium  molybdate,  0.75  c.c.  was  found  necessary 
to  turn  the  indicator :  0.300  gram  of  lead  took  32  c.c.  of 
the  solution  (without  allowance).  If  the  subtraction  of 
the  blank  is  omitted,  what  is  the  error  in  the  titration  of 
one  gram  of  a  60%  ore  ?  Ans.     0.7 1%. 

19.  An  ammonium  molybdate  solution  was  standard- 
ized by  lead  sulphate,  without  making  any  allowance  for 
the  end  point ;  40  c.c.  were  required  for  0.55   gram  of 


PRECIPITATION.  69 

lead  sulphate.  A  lead  ore  was  titrated  with  this  solution, 
0.5  gram  required  12  c.c.  If  the  proper  allowance  for 
the  end  point  were  0.7  c.c,  what  is  the  true  percentage 
of  lead  ?  Ans,     2 1 .  60%. 

20.  If  the  proper  allowance  for  turning  the  indicator 
is  0.2  c.c.  in  titradng  chlorine  with  ^j-^^  AgNOg,  what 
difference  will  the  omission  of  this  allowance  make  in 
titrating  i  litre  of  water  containing  20  parts  per  million 
of  chlorine  ?  Ans.     .07 1  parts  CI  per  million. 


PART   II.     REACTIONS   OF   PRECIPITATION  AND   OF 
SATURATION. 

PRECIPITATION. 

Usually  in  all  cases  where  either  normal  or  standard 
solutions  are  used  for  precipitation  the  reaction  is  simple 
and  the  precipitate  of  definite  composition,  so  that  these 
calculations  require  no  additional  explanation.  The  only 
cases  likely  to  give  trouble  are  those  in  which  a  reagent 
combines  in  different  and  often  complex  ratios  with  differ- 
ent metals,  so  that  reasoning  by  analogy  without  a 
knowledge  of  the  exact  composition  of  the  precipitate 
leads  to  error. 

For  example,  what  would  be  the  strength  of  a  normal 
potassium  ferrocyanide  solution  for  precipitating  zinc? 
The  reaction  in  an  acid  solution  and  in  the  absence  of 
ammonium  chloride  is, 

3ZnCl2  +  2K4Fe(CN)e  =  K2Zn3(Fe(CN)e)2  +  6KC1 ; 
so  two  molecules  of  potassium  ferrocyanide  bring  into 
reaction  three  atoms  of  zinc,  each  equivalent  to  two  of 
hydrogen.  The  normal  solution  is  therefore  the  molec- 
ular weight  in  grams  divided  by  three  or   122.91  grams. 


70  ALKALIMETRY   AND   ACIDIMETRY. 

To  get  this  amount  of  the  active  reagent  we  must 
weigh  out  140.93  grams  of  the  crystalHzed  salt  as  it  con- 
tains three  molecules  of  water. 

How  much  K4Fe(CN)6,  3H2O  must  be  dissolved 
and  diluted  to  one  litre  to  give  a  solution  of  such  strength 
that  I  c.c.  =  0.005  gram  of  zinc  ? 

The  molecular  weight  is  422.79.     As  two  molecules 
precipitate  three  atoms  of  zinc  we  have  the  proportion 
845.58  :  196.2  : :  X  :  5.     x  =  21.549. 

Ferrocyanide  solutions  are  also  used  for  the  precipita- 
tion of  lead  and  of  cadmium,  but  the  formulae  of  the  pre- 
cipitates are  approximately 

Pb2Fe(CN)6  and  K2CdFe(CN)6, 
so  that  all  calculations  based  on  simple  ratios  between 
the  atomic  weights  will  be  wrong  and  the  reaction  in  each 
case  must  be  made  the  basis  for  the  computation. 

In  cases,  like  the  ferrocyanides,  where  the  composition 
of  the  precipitate  varies  with  varying  conditions  in  the 
titration,  there  is  no  advantage  in  using  normal  solutions. 

ALKALIMETRY    AND    ACIDIMETRY. 

Many  of  these  calculations  require  no  special  men- 
tion here  as  they  can  readily  be  performed  after  the 
explanations  already  given.  Only  a  few  special  cases 
are  given  in  detail. 

In  order  not  to  complicate  these  calculations  by  the 
use  of  different  indicators,  phenolphthalein  is  used 
throughout.     This  indicator  acts  as  follows  : 

Red  with  fixed  alkalies,  colorless  with  acids,  includ- 
ing carbonic.  So  when  acid  is  run  into  carbonated 
alkali  the  color  is  discharged  when  the  NasCOg  is  con- 
verted into  NaHCOa,  but  on  boiling  the  color  returns 


ALKALIMETRY   AND   ACIDIMETRY.  71 

due  to  the  formation  of  more  NagCOa  from  the  NaHCOa, 
so  to  get  total  NagCOs,  acid  must  be  run  in  till  the  solu- 
tion remains  colorless  on  boiling.     The  reactions  are 
NaoCOa  +  HCl  =  NaHCOs  +  NaCl  and  (boiling) 
NaHC03  +  HCl  =  NaCl  +  CO^  +  H^O. 

I.  Caustic  and  Carbonated  Alkali. 

1.5  grams  of  impure  caustic  potash  are  dissolved  in 
water,  diluted,  an  excess  of  barium  chloride  added 
and  diluted  to  400  c.c.  One  hundred  cubic  centimeters 
are  filtered  off  and  titrated  with  N  HCl,  5.4  c.c.  required. 
Another  i .  5  grams  of  the  sample  dissolved  and  titrated 
directly  with  N  HCl  boiling,  for  total  alkalinity,  re- 
quired 24.6  c.c.  What  are  the  weights  of  KOH  and 
ofKgCOg?      The  first  titration  gives  the  caustic  alkaU 

in  — —  grams,   for  the  barium  chloride  precipitates  the 

carbonate  as  barium  carbonate  and  with  the  caustic  gives 
equivalent  amount  of  Ba  (OH)2,  which  is  determined  by 
the  N  HCl.     The  reactions  are 

Na^COo  4-  BaCls  =  BaCOa  +  2NaCl  and 
2NaOH  +  BaClo  =  Ba(OH)2  +  2NaCl 
The  total  required  24.6  c.c. 

The  caustic  required      21.6  c.c. 
The  carbonate  required  3.0  c.  c 
N  HCl  =  0.056158  gram  KOH  per  c.c. 
=  0.06915  gram  KgCOg  per  c.c. 
hence  the  weights  are  KOH,  1.2 130  grams  and  Kg  CO  3 
0.20745  gram. 

2.  Carbonate  and  Bicarbonate. 

Four  grams  of  commercial  sodium  bicarbonate  are 
titrated  in  a  cold  dilute  solution  with  %  H0SO4  till  the 
red  color  disappears  :  used  6.4  c.c. 


72  ALKALIMETRY   AND   ACIDIMETRY. 

To  one  gram  of  the  sample  dissolved  in  water,  20  c.c. 
of  N  H0SO4  are  added  and  the  solution  boiled  till  all 
the  carbonic  acid  is  expelled,  then  titrated  back  with 
N  KOH  :  requiring  8.1  c.c. 

What  are  the  percentages  of  NaHCOg  and  of 
NaXOs? 

The  first  titration  shows  one  half  of  the  Na2C03,  so 
for  one  gram  the  Na2C03  required 

L^L^  =  3.2  c.c.  of  %  HoSO,  or  0.32  c.c.  N  H2SO4. 

4 

The  total  NasO  required  20-8.1  =  1 1.9  c.c.  N  H0SO4 
so  the  N  H2SO4  used  on  the  NagO  present  as  NaHCOg 
was  1 1.9  —  0.32  =  1 1.58  c.c. 

The  weights  corresponding  are 

NaHCOg  1 1.58  X  0.084058  =  0.9734  gram 
NasCOg    0.32  X  0.05305    =  0.0169  gram 
or  97.34%  NaHCOg  and  1.70%  Na^COg. 

3.   Phosphorus  from  Ammonium  Phospho-Molybdate. 

The  phosphorus  is  precipitated  as(NH4)3P04,  12M0O3, 
under  carefully  regulated  conditions,  this  is  dissolved 
in  a  standard  alkali  solution  according  to  the  reaction  ; 

2  (NH4)3P04, 1 2M0O3  +  23NaoC03  +  HoO= 
2(NH4)2HP04+(NH4)2Mo64+23NaoMo04+23COo. 

In  practice  an  excess  of  the  standard  Na2C03  solu- 
tion is  added  then  the  excess  titrated  using  phenol- 
phtalein  as  an  indicator.  The  reason  that  two  molecules 
of  the  precipitate  require  23  and  not  24  molecules  of  Nag 
COg  is  that  this  indicator  changes  color  when  two  of  the 
three  replaceable  hydrogens  of  phosphoric  acid  have 
been  substituted  by  alkali,  which  leaves  (from  two  mole- 
cules) two  NH4  groups  to  combine  with  a  molecule  of 


ALKALIMETRY   AND   ACIDIMETRY.  73 

M0O3  thus  leaving  23  instead  of  24  molecules  of  M0O3 
to  be  neutralized  by  NaoCOs. 

To  calculate  the  grams  per  liter  of  NasCOa  solution 
equivalent  to  one  milligram  of  P2O5  per  cubic  centi- 
meter, we  have:   i.c.c.  =  i  mg.     i  liter  =  i  gram. 
2(NH4)3P04,  12M0O3   contain    iP^Og  hence 

Mol.  wt.  P2O5  :    23(Mol.wt.Na2C03)  :  :  i  :  x 
142.  :  2440.3  :  :  I  :  X 

x=  1 7. 1 85  grams  per  liter  of  NagCOa. 

4c  Sulphuric  and  Nitric  Acids  in  Mixed  Acid. 

The  analysis  of  the  waste  mixed  acid  from  nitrating 
IS  sometimes  made  as  follows  :  50  grams  of  the  sample 
are  diluted  to  a  liter  and  10.  c.c.  of  this  mixture  are  with- 
drawn by  a  pipette  which  in  this  case  delivered  an  amount 
containing  exactly  0.4976  gram  of  the  original  sample. 
The  sulphuric  acid  in  this  is  precipitated  as  BaSO^  and 
weighed  =  o.  7093  gram. 

Another  portion  is  titrated  for  total  acidity  as  follows: 
Of  an  approximately  N  NaOH  solution  (20.2  c  c.  =  0.98 
gram  1^0504)20.2  c.c.  are  run  into  a  beaker  and  this  is 
the  neutralized  by  the  diluted  acid  solution.  (50  grams 
to  liter.)     Suppose  25.75  c.c.  are  required. 

To  calculate  the  percentage  of  H0SO4 

0.7093X0.4201  X  joo  ^  59.88% 
0.4976 
To  calculate  the  percentage  of  HNO3 
25.75  c.c.  =  2o.2  c.c.  NaOH  =  0.98  gram  H0SO4 

First  calculate  the  total  acidity  to  H0SO4 : 
As  50  grams  are  contained  in   1000  c.c.    25.75  c.c. 
contain  1.2875  grams  for  50:    1000:  :  x  :  25.75     or  x  = 

^^•^5=  1.2875  grams.     Hence  the  total  acidity  in  terms 
20 


74  ALKALIMETRY   AND   ACIDIMETRY. 

of  HoSO^is-^:^  X  loo  or  76.12%  and  76.12%  - 

1.2875  '  /u  /  /u 

59.88%  or  16.24%  is  the  acidity  due  to  nitric  acid  but 
expressed  in  terms  of  H2SO4.  As  63.048  parts  of  HNO3 
are  equivalent  in  power  of  neutralizing  NaOH  to  49.043 
parts  of  H2SO4 ;  to  determine  the  percentage  of  HNO3 
we  have  :  16.24  :  x  :  :  49.038  :  63.048.  x  =  20.88%. 
So  the  composition  of  the  waste  acid  is 

H2SO4    59-88%,  HNO3  20.88%. 


EXAMPLES. 

1.  How  many  c.c,  more  of  ^Vio  AgNOg  will  it  take  to 
precipitate  2  grams  of  KCl  than  2  grams  of  KI  ? 

Ans.     147.6  c.c. 

2.  How  much  more  ^/^  K2Cr04  than  ^/^  Na2Cr04 
will  it  take  to  precipitate  i  gram  of  BaClg  ? 

Prove  that  it  takes  same  amount. 

3.  One  gram  of  an  ore  containing  arsenic  was  treated 
according  to  Pearce's  method  and  the  arsenic  precipitated 
as  Ag3As04  ;  the  silver  so  combined  was  titrated  by 
NH4SCN.  The  amount  used  was  27.9  c.c.  and  the 
strength  of  the  solution  i  c.c.  =  0.007  gi*am  of  silver. 
What  was  the  percentage  of  arsenic  ? 

Ans.     4.52%. 

4.  How  much  crystallized  potassium  ferrocyanide  per 
liter  will  give  a  solution  of  such  strength  that  each  c.c.  is 
equal  to  1%  of  lead  when  0.5  gram  of  ore  is  taken  for 
analysis  ?  Given  the  composition  of  the  precipitate  as 
Pb2Fe(CN)6  Ans.     5.1086  grams. 

5.  How  many  grams  more  of  lead  than  of  cadmium 
will  be  precipitated  by  25  c.c.  of  ^/lo  potassium  ferro- 
cyanide solution  (for  zinc)  given  the  precipitate  in  the 
case  of  cadmium  as  K2CdFe(CN)6  ? 

Ans.     0.25 1 16  gram. 

6.  How  much  more  N  NaOH  will  it  take  to  neu- 
tralize I  gram  of  HCl  than  i  gram  HBr? 

Ans.     15.08  c.c. 

(75) 


76  EXAMPLES. 

7.  How  much  more  N  NaOH  will  it  take  to  neu- 
tralize I  gram  of  NaHS04  than  N  KOH  to  neutralize  i 
gramofKHS04?  Ans.     0.99  c.c. 

8.  How  much  sodium  hydroxide  must  be  added  to 
seven  liters  of  HCl  i  c.c.  =0.04  gram  HCl  to  make  it 
normal?  Ans.     27.2422  grams. 

9.  How  many  grams  of  20%  HCl  must  be  added  to 
five  liters  of  potassium  hydroxide  solution,  i  c.c.  =  0.004 
gram  of  HCl  to  make  it  ^/lo  ?  Ans.     8.855  grams. 

10.  How  much  more  ^/lo  HBr  will  it  take  to  neu- 
tralize one  gram  of  NaOH  than  it  does  of  ^/lo  HCl  to 
neutralize  i  gram  of  KOH  ?  Ans.     71.5  c.c. 

11.  Given  a  solution  of  sodium  hydroxide  containing 
exactly  30  grams  per  liter.  How  much  ^/^q  NaOH  must 
be  added  to  one  liter  to  make  a  ^/g  solution  ?  .  How 
much  ^72  H2SO4  must  be  added  to  a  second  liter  to 
make  a  ^/lo  solution  ?     Ans.     5489  c.c.        108 1.5  c.c. 

12.  What  is  the  value  of  N  NaOH  in  terms  of  phos- 
phorus from  ammonium  phosphomolybdate  ? 

Ans.     0.0013478  gram. 

13.  How  much  water  must  be  added  to  i  liter  of  ^/^o 
NaOH  so  that  each  c.c.  will  read  0.01%  of  phosphorus 
when  I  gram  of  steel  is  taken  for  analysis  ? 

Ans.     347.8  c.c. 

14.  Two  liters  of  ^/lo  barium  hydroxide  solution  be- 
came cloudy  by  absorption  of  carbon  dioxide  ;  it  was 
filtered  in  air  free  from  CO  2,  and  the  barium  carbonate 
thoroughly  washed.  The  volume  of  the  barium  hydroxide 
solution  with  washings  was  2300  c.c.  The  weight  of 
BaCOg  was  270  milligrams.  How  much  barium  hy- 
droxide must  be  added  to  make  the  solution  ^7 10  again  ? 

Ans.     2.8058  grams. 


EXAMPLES.  77 

15.  0.2  gram  of  a  nitrogenous  organic  compound  was 
heated  with  soda  Hme  and  the  NH3  evolved  caught  in 
50  c.c.  ^/lo  HCl :  the  excess  of  HCl  was  neutraHzed  by 
14  c.c.  of  ^Vs  NaOH.  What  was  the  percentage  of 
nitrogen?  Ans.     15.44%. 

16.  Twenty  grams  of  an  adulterated  sample  of  vin- 
egar, freed  from  soluble  sulphates,  required  16  c.c.  N 
NaOH  :  10  grams  gave  0.3162  gram  of  BaS04.  What 
is  the  percentage  of  acetic  acid  ?  Ans.     3.177%. 

17.  0.4  gram  of  impure  caustic  soda  required  when 
titrated  directly  95  c.c.  of  ^/lo  HCl:  another  0.4  gram 
was  dissolved,  an  excess  of  barium  chloride  added,  and 
then  one  half  titrated,  requiring  46  c.c.  What  are  the 
percentages  of  NaOH  and  Na2C03  ? 

Ans      "1  NaOH  92.13%. 
^'^"-      t  Na.COs  3-98%. 

18.  Two  grams  of  impure  potassium  bicarbonate  re- 
quired when  titrated  cold,  in  a  dilute  solution,  6  c.c.  of 
^/lo  H2SO4  :  two  grams  when  titrated  boiling  hot  re- 
quired 18  c.c.  of  N  H2SO4.  What  are  the  percentages 
ofK^COs  andofKHCOg.? 

Ans.     4.15%  and  84.13%. 

19.  Two  grams  of  impure  caustic  soda  were  dissolved 
in  water,  an  excess  of  barium  chloride  added  and  diluted 
to  500  c.c.  ;  100  c.c.  was  titrated  with  ^72  HCl  and  re- 
quired 1 7  c.c.  Another  two  grams  were  titrated  also  with 
^/g  HCl,  boiling,  for  total  alkalinity;  93  c.c.  were  re- 
quired. What  are  the  percentages  of  NaOH  and  of 
Na2C03?  Ans.     ^.i  2%,  10.61%. 

20.  Three  grams  of  commercial  sodium  bicarbonate 
were  titrated  with  ^/s  HCl  in  a  cold  dilute  solution  till 
the  red  color  disappeared,  4.2  c.c.  were  required.     One 


78  PERMANGANATE  CALCULATIONS. 

gram  of  the  sample  was  dissolved  in  water,  diluted  largely 
and  30  c.c.  of  ^/g  HCl  added,  then  boiled  till  all  CO2  was 
expelled,  and  the  excess  of  acid  titrated  by  ^/a  KOH  re- 
quiring 6.2  c.c.  What  are  the  percentages  of  Nag  CO  3 
and  NaHCOg?  Ans.     2.97%,  and  95.32%. 

PART  III.     REACTIONS  OF  OXIDATION  AND  REDUCTION. 
PERMANGANATE  CALCULATIONS. 

Suppose  we  have  a  ^/g  solution  of  KMn04 ;  to  cal- 
culate its  streno-th  in  terms  of  oxalic  acid  we  have  the  re- 
actions, 
2KMn04  -r  ioFeS04  +  8H2SO4  = 

K2SO4  +  2MnS04  +  5Fe2(S04)3  +  8H2O 
2KMn04  +  5H2C2O4  +  3H2SO4  = 

K2SO4  +  2MnS04  +  10CO2  +  8H2O. 

We  see  that  the  same  quantity  of  KMn04  is  required 
for  ioFeS04  as  for  5H2C2O4;  ioFeS04  contains  10 
Fe  so  that  as  regards  KMn04,  10  X  atomic  wt.  of  Fe  = 
5  X  Mol.  wt.  H2C2O4  or  we  have,  Fe  standard  o.oi  118  : 
oxalic  acid  standard  ::ioX55.9:5X  90.016  or 

0.01 1 18  :  X  :  :  559.  :  450.08.     x  =0.009002. 

All  these  steps  are  not  necessary  however,  all  we 
need  do  is  to  consider  the  strength  of  ^/g  oxalic  acid, 
then  this  will  be  equivalent  c.c.  per  c.c.  to  ^/s  KMn04. 
From  the  reaction  we  see  that  one  H2C2O4  requires  one 
O  :  therefore  the  amount  necessary  to  bring  into  reaction 
eight  grams  of  oxygen  will  be  one  half  the  molecular 
weight  in  grams  per  liter  and  for  ^/g  one  tenth  the  molec- 
ular weight  or  9.0016  grams,  so  that  i  c.c.  will  be  equal 
to  0.009002  gram. 


PERMANGANATE  CALCULATIONS.  79 

Many  metals  are  precipitated  as  normal  oxalates  such 
as  calcium,  barium,  lead,  cadmium,  etc.  (not  bismuth). 
In  such  cases  the  metal  combined  with  the  oxalic  acid 
can  be  determined  indirectly  by  decomposing  the  pre- 
cipitate with  sulphuric  acid  and  titrating  the  oxalic  acid 
set  free.  To  calculate  the  standard  of  the  permanganate 
solution  we  need  only  remember  that  i  CaO  or  i  Pb  cor- 
responds to  one  oxalic  acid.  So  as  ^/^  oxalic  acid  was 
one  tenth  the  molecular  weight  in  grams,  ^/g  CaO  will 
also  be  one  tenth,  or  5.61  grams,  or  ^/g  KMn04  is 
equivalent  to  0.00561  gram  of  CaO  per  c.c. ;  very  nearly 
one  half  the  iron  standard. 

To  calculate  the  strength  of  ^/g  KMn04  in  terms  of 
molybdic  oxide  we  have  the  oxidation  of  M02O3  to 
M0O3  or  two  M0O3  require  3  oxygen,  and  we  have 

288  :  x  :  :  48  :  8.     x  =  48. 

Therefore  N  M0O3  would  have  one  third  the  molecular 
weight  in  grams  per  liter  (48  grams)  and  ^/g  one  fif- 
teenth (9.6  grams)  per  liter  or  i  c.c.  =  0.0096  gram 
M0O3. 

The  determination  of  phosphorus  by  KMn04  is  in- 
direct for  as  the  ratio  of  M0O3  to  P  in  the  ammonium 
phospho-molybdate  is  12  to  i,  for  every  144  X  12  grams 
of  M0O3  found  by  titration  there  must  be  31  grams  of 
phosphorus  present  :  so  we  have  M0O3  standard  :  P 
standard  as  1728  :  31  or  i  c.c.  ^/g  KMn04=  0.0001722 
gram  of  phosphorus. 

To  calculate  this  value  for  a  standard  KMn04  solution 
not  normal  when  the  iron  standard  is  known  :  we  have 

6KMn04  =  10M0O3  and 

6KMn04  =  30  Fe  hence 


so  PERMANGANATE   CALCULATIONS. 

10M0O3  correspond  to  3oFe  and  the  Fe  standard  : 
M0O3  standard  :  :  1677  :  1440  and  as  the  phosphorus 
standard  is  1.794%  of  the  M0O3  standard,  the  phos- 
phorus standard  will  be  1.54%  of  the  Fe  standard  or  the 
Fe  standard  times  0.01540. 

The  degree  of  reduction  obtained  in  a  reductor  is  not 
that  just  given  but  corresponds  very  closely  to  M024O37, 
so  that  the  molybdenum  and  phosphorus  standards  are 
slightly  higher  than  when  the  complete  reduction  to 
MogOg  is  effected. 

To  calculate  these  values  for  ^/jq  KM  n04  we  have 
I  M024O37  requires  35  oxygen  equivalent  to  70  iron 
14  KMn04  give  35  oxygen  equivalent  to  70  iron 
^/lo  KMn04  =  1/50  mol.  weight  in  grams  per  litre 

^/lo  M024O37  = — — mol.  weight  in  grams  per  litre 

so  to  calculate  the  molybdenum  standard  of  ^/^o  KMn04 
per  c.c.  we  have 

— -^ -^ or  0.00^2014  orram  molybdenum. 

I4X5OX    1000  O      y     t    £>  ^ 

To  derive  this  from  the  iron  standard,  we  have  24  Mo 
equivalent  to  7oFe,  hence 

24X96  •  70X55-9  ••  X  :  0.00559.     X  =0.0032914. 
To  obtain  the   phosphorus  standard,  we  have  from 
the  ratio  of  molybdenum  to  phosphorus  in  the  ''yellow 
precipitate  " 

1 2  X  96  :  31  : :  Mo  standard  :  P  standard 
or  1 152  :  31  : :  0.0032914  :  y. 

y  =  0.00008857  gram  phosphorus  per  c.c. 
A  rather  confusing  calculation  is  met  in  the  Ford- 
Williams  method  for  manganese :  here  the  manganese 


PERMANGANATE   CALCULATIONS.  81 

IS  precipitated  as  hydrated  MnOg  and  its  oxidizing 
power  measured.  Suppose  we  take  for  analysis  one 
gram  of  pig  iron,  and  to  the  precipitate  of  MnOg  add 
25  c.c.  of  ^/lo  FeS04  and  then  titrate  the  FeS04  remain- 
ing unoxidized  by  ^/lo  KMn04,  using  12  c.c.  What  is 
the  percentage  of  manganese? 
The  reactions  are 

MnOa  +  2FeS04  +  2H2SO4  = 

MnS04  +  Fe2(S04)3  +2H2O. 

and  the  usual  reaction  between  the  excess  of  ferrous 
sulphate  and  permanganate.  The  solutions  used  were 
both  ^/lo  and  therefore  correspond  c.c.  for  c.c,  so  that  the 
oxidation  done  by  the  Mn02  was  equal  to  that  which 
would  have  been  done  by  13  c.c.  of  ^/lo  KMn04.  Now 
we  have 

2KMn04  =  loFe, 

iMnOs      =    2Fe. 

Hence  2KMn04  are  equivalent  to  sMnOg  or  to  sMn 
and  we  have  Fe  standard  :  Mn  standard  : :  10  times  at. 
wt.  Fe  :  5  times  at.  wt.  Mn  or  0.00559  :  x  :  :  559.  :  275 
x  =  0.00275.  The  percentage  of  manganese  will  be  13 
(number  of  c.c.  of  KMn04  no^  required) 

X  0.00275  ^  I  X  100  or  3.575%  Mn. 

Oxalic  acid  may  be  used  in  the  same  method  instead 
of  ferrous  sulphate  to  measure  the  oxidizing  power  of  the 
precipitate  and  the  excess  titrated  by  permanganate. 
Suppose  one  gram  of  spiegel  is  taken  for  analysis  and 
that  we  have  a  KMn04  solution  (not  systematic)  whose 
strength  is  i  c.c.  =  0.006  gram  of  iron.  We  make  up 
a  solution  of  oxalic  acid  and  find  that  it  requires  25  c.c. 


82  PERMANGANATE   CALCULATIONS. 

of  the  KMn04  solution  to  oxidize  20  c.c.  of  the  oxalic 
acid  solution.  To  the  precipitate  (hydrated  MnOa) 
from  one  gram  of  spiegel,  80  c.c.  of  the  oxalic  acid  solu- 
tion are  added  and,  after  the  complete  solution  of  the  pre- 
cipitate, the  excess  is  titrated  by  KMn04,  this  requires 
12  c.c.  To  calculate  the  percentage  of  manganese.  If 
there  had  been  no  MnOg  present  the  80  c.c.  of  oxalic 
acid  would  have  required  100  c.c.  of  KMn04,  but  only 
12  were  needed,  therefore  the  oxidation  done  by  the 
MnOg  present  was  equal  to  that  of  88  c.c.  of  KMn04. 
Now  the  manganese  standard  of  our  KMn04  is 

0.006  :  X  : :  559  :  275  or  0.002952 
and 

88  X  0.002952  ^  I  X  100  =  25.98 

the  percentage  of  manganese. 

The  peculiar  part  of  this  method  Is  that  it  is  the 
amount  of  KMn04  not  used  which  is  the  basis  of  calcu- 
lation. 

The  problem  of  making  a  potassium  dichromate 
solution  equivalent  to  an  unsystematic  permanganate 
solution  sometimes  gives  trouble.  Suppose  we  wish  to 
make  up  two  liters  of  dichromate  equivalent  in  oxidizing 
power  to  a  permanganate  solution  containing  three  grams 
per  liter,  how  much  K2Cr207  is  to  be  weighed  out? 
The  molecular  weight  of  KgCrgOy  is  294.5  ;  294.5  grams 
oxidize  6  X  55-9  grams  of  iron. 

The  molecular  weight  of  KMn04  is  158.15;  158.15 
grams  oxidize  5  X  55-9  grams  of  iron. 

We  must  calculate  the  weight  of  KgCrgOy  equal  in 
oxidizing  power  to  six  grams  of  KMn04  : 

1 158.  IS     294.5  o 

6:x::-5 ^  ;  _^^.     x  =  9.3108. 

50 


IODINE   CALCULATIONS.  83 

IODINE    CALCULATIONS. 

Tenth  normal  iodine  solution,  12.685  grams  per  liter, 
can  be  used  for  a  great  many  titrations.  For  example, 
in  the  determination  of  HoS  and  so  of  sulphur  in  pig 
iron,  &c.     The  reaction  is 

HoS  +  2l  =  S+2HI. 
Five  grams  of  pig  iron  evolved,  on  treatment  with 
hydrochloric  acid,  HoS,  which  when  titrated  in  a  dilute 
solution  with  starch  as  indicator  required  6.2  c.c.  of  ^/^q 
iodine  solution.     What  is  the  percentage  of  sulphur  ? 
^/lo  I  =  0.012685  gram  I  per  c.c.  =  0.001603 
gram  S  per  c.c.         6.2  x  0.0016  ^  ^^^  ^  0.198%  S. 

Iodine  reacts  with  H0SO3  as  follows : 

I2  +  H0SO3  +  HoO  =  2HI  +  H2SO4 
and  with  NagSoOs  according  to  the  reaction, 
I2  +  2NaoS203  =  2NaI  +  NaoS406 

we  see  therefore  that  one  iodine  is  equivalent  to  }4  SO2, 
}i  H2SO3  or  }4  Na2S03  and  to  one  Na2S203. 

The  latter  reaction  can  be  employed  for  the  deter- 
mination of  copper  as  follows : 

2CU  (C2H302)o  +  4KI  =  CU2I2  +  I2  +  4KC2H3O2 

and  taking  the  last  two  reactions  we  see  that  one  mole- 
cule of  Na2S203  equivalent  to  one  iodine  is  equivalent 
to  one  copper,  so  if  we  have  a  ^/lo  iodine  solution  we  can 
use  it  to  standardize  a  sodium  thiosulphate  solution  and 
so  indirectly  determine  copper  by  measuring  the  iodine 
liberated. 

Suppose  we  have  an  iodine  solution  of  unknown 
strength  and  a  potassium  dichromate  solution  whose 
strength  is  i  c.c.  =  0.00559  gram  of  iron.  To  find  the 
strength  of  our  iodine  solution  in  terms  of  sulphur. 


84  IODINE   CALCULATIONS. 

Make  up  a  sodium  thiosulphate  solution  of  any  con- 
venient strength  say  24.8  grams  of  the  crystalUzed  salt 
(NaoSoOg.  5H0O)  per  liter,  approximately  ^7io- 

Dissolve  one  gram  of  potassium  iodide  in  water, 
acidify  strongly  with  HCl  and  add  25  c.c.  of  the  dichro- 
mate  solution.  Iodine  will  be  liberated  according  to  the 
reaction 

K2Cr207  +  6KI+i4HCl  =  3lo+8KCl+CroCl6+7HoO 
or  the  amount  of  KoCr207  which  oxidizes  6Fe  liberates 
61 ;  therefore  the  strength  of  the  KoCroO;  solution  is 
I. c.c.  =  0.012685  gram  of  iodine  and  the  weight  of 
iodine  liberated  will  be  25  X  o.oi  2685    ^^  0.3 1 7 1  gram. 

Into  this  we  run  the  sodium  thiosulphate  solution  till 

it  gives  no  blue  color  with  starch  ;  suppose  this  requires 

25.5  c.c.  then   i.c.c.  =  0.01243   gram  of  iodine.     Now 

take   any  convenient   amount   of  the  unknown   iodine 

solution    as    25    c.c.    and   titrate    it    with   the   sodium 

thiosulphate  solution;  suppose  it  requires   12.2  c.c.  then 

.         ^         ^1     .     ^               r   •   J-        .      12.2  X  0.01243 
Its   strenerth    m  terms    01    lodme    is     —   or 

0.006066  and  against  sulphur  when  titrated  as  HgS, 
0.006066  :  X  : :  126.85  •  ^^-^3 
X  or  I  c.c.  =  0.000767  gram  sulphur. 
Arsenic   is  determined   by  iodine  according  to  the 
reaction : 

NagAsOg  +  I2  +  2NaHC03  = 
Na3As04  +  2NaI  +  H2O  +  2CO2 
from  which  it  is  evident  that  2I  =  lAs  or  i  c.c.  ^/iq 
I  =  0.00375  gram  of  arserfic  or  0.00495  gi*am  of  AS4O6. 
The  available  chlorine  in  bleaching  powder  can  be  de- 
termined according  to  a  similar  reaction  by  using  a  solu- 
tion of  NagAsOg,  which  can  conveniently  be  standard- 
ized by  the  ^/lo  iodine. 


IODINE   CALCULATIONS.  85 

Antimony  may  be  determined  by  sodium  thiosulphate 
in  a  hydrochloric  acid  solution  as  follows : 

HsSbO^  +  5KI  +  5HCI  =  Sbl3  +  I.  +  5KCI  +  4H2O 
lo  +  2Na2S203  =  2NaI  +  Na2S406. 

Suppose  we  have  a  solution  of  NagSgOg  whose 
strength  is  i  c.c.  =  0.005  gram  Cu.  What  is  its  strength 
in  terms  of  antimony  ? 

We  see  by  comparing  the  reactions  that  2Cu  =  1 2  and 
iSb  =  1 2  hence 

0.005  •  X  : :  63.6  :  60.1.     x  =  0.00472  gram. 

If  the  sodium  thiosulphate  were  ^/jq  we  should  see  at 
once  its  strength  against  copper  and  antimony,  for  one 
atom  of  copper  brings  into  reaction  one  iodine  equivalent 
to  one  half  atom  of  oxygen  or  a  ^/lo  solution  of  Cu  = 
6.36  grams  per  liter ;  while  i  Sb  =  2I  =  i  oxygen.  So 
^VioSb  =  120.2  ^  (2  X  10)  =  6.01  grams  and 
I  c.c.  ^/lo  NagSsOg  =  0.00636  gram  Cu 
=  0.00601  gram  Sb. 
Similarly,  but  in  a  more  strongly  acid  solution  arsenic 
acid,  H3ASO4  liberates  iodine  being  itself  reduced  to  the 
arsenious  condition.  The  strength  of  ^/jo  sodium  thio- 
sulphate would  be  I  c.c.  =  0.00375  gram  of  arsenic. 
When  antimony  is  oxidized  from  the  triad  to  the  pentad 
condition  in  a  solution  alkaline  with  bicarbonate  the  cal- 
culation is  the  same  as  that  already  given  for  arsenic,  so 
that  each  c.c.  of  ^/jo  iodine  is  equivalent  to  0.00601  gram 
of  antimony. 

REDUCTION    BY   STANDARD  SOLUTIONS. 

How  many  grams  of  tin  must  be  dissolved  in  hydro- 
chloric acid  and  diluted  to  one  liter,  to  give  a  solution  of 


86  REDUCTION  BY  STANDARD   SOLUTIONS. 

such  strength  that  each  c.c.  will  be  equivalent  to  i%  of 
iron  when 0.5  gram  is  taken  for  analysis? 

In  order  to  read  percentage  from  the  burette,  each 
c.c.  must  reduce  0.005  gram  of  iron,  each  litre  5  grams  : 
from  the  reaction,  SnClg  +  2FeCl3  =  SnCl4  +  2FeCl2, 
we  see  that  one  molecule  of  stannous  chloride  containing 
one  atom  of  tin  reduces  two  atoms  of  iron ;  so  the  pro- 
portion is 

5  :  X  ::  55.9  X  2  :  119.     x  =  5.3220. 

How  many  grams  per  liter  of  Mohr's  salt  (NH 4)2804, 
FeS04,  6H2O,  will  give  a  solution  equivalent  to  ^/lo 
K2Cr2  07  ?  As  all  ^/^q  solutions  are  equivalent  the  prob- 
lem is  to  calculate  the  number  of  grams  for  a  ^/jq  solution 
of  Mohr's  salt.  One  molecule  of  Mohr  s  salt  contains 
one  FeO,  which  requires  for  oxidation  half  an  atom  of 
oxygen,  equivalent  to  one  atom  of  hydrogen,  hence  the 
normal  solution  is  the  molecular  weight  in  grams,  and  the 
^Jio  39.226  grams  per  liter. 


EXAMPLES. 

1.  Calculate  the  strength  of  ^/go  KMn04  to  be  used 
in  titrating  K4Fe(CN)6. 

Ans,     1.5815  grams  per  liter. 

2.  What  is  the  strength  of  ^/lo  KMn04  in  terms  of 
Fe?  Fe203?  Fe.^O^  ?  FeS04  7H2O? 

Ans.     0.00559,  0.00799,  0-007723,  0.02781. 

3.  What  is  the  strength  of  ^/lo  KMn04  in  terms  of 
H2C2O4?  ofCaO?  ofCaCOg?  of  CaS04  ? 

Ans.     0.0045.     0.002805.     0.005005.     0.006808. 

4.  What  is  the  strength  of  ^/lo  KMn04  in  terms  of 
P  and  of  M0O3  according  to  Noyes  method? 

Ans.     0.0000861  and  0.0048. 

5.  1.5  grams  of  pig  iron  took  y^  c.c.  of  ^/lo  KMn04. 
What  is  the  percentage  of  phosphorus  ? 

Ans.     0.447%. 

6.  In  determining  iron  in  an  ore,  by  mistake  too 
much  ^/lo  K 2^207  was  run  in  :  the  burette  read  45  c.c: 
20  c.c.  of  ^/lo  FeS04  were  added  and  the  titration 
finished:  the  burette  then  read  57  c.c.  How  much  di- 
chromate  was  required  for  the  iron  ore?  Ans.     2>7  ^-C- 

7.  Given  a  solution  of  KMn04  i  c.c.  =  0.00559 
gram  of  iron.  What  is  its  strength  in  terms  of  manga- 
nese by  the  Volhard  method  and  by  the  Ford-Williams 
method  ? 

r  Volhard  i  c.c.  =  0.00165  gram. 

(  Ford-Williams  i  c.c.  =  0.00275  gram. 

(87) 


nS  EXAMPLES. 

8.  The  MnOg  from  one  gram  of  spiegel  was  dis- 
solved in  1.5  grams  of  ferrous  ammonium  sulphate, 
(NH4)2S04,  FeS04,  6H2O,  and  then  titrated  by  ^/lo 
KMn04  :  2.05  c.c.  were  used.  What  is  the  percentage 
of  manganese?  Ans.     9.95%. 

9.  3  grams  of  steel  required  60  c.c.  of  ^/lo  KMn04  to 
oxidize  the  molybdenum,  from  the  ammonium  phospho- 
molybdate  precipitate,  from  M024O37  to  M0O3.  What 
is  the  percentage  of  phosphorus  ?  Ans.     .177% 

10.  A  solution  of  Na2S2  03  was  made  containing 
about  40  grams  of  the  crystallized  salt  per  liter  ;  this  was 
standardized  by  portions  of  copper  0.200  gram  each  ; 
they  required  19.6  c.c.  :  0.50  gram  of  a  copper  matte 
took  37.6  c.c.     What  is  the  percentage  of  copper? 

Ans,     7^'73%- 

1 1 .  Calculate  the  number  ot  grams  per  liter  to  give 
half  normal  solutions  of  I,  of  Na2S2  03,  sHoO,  and  of 
SO2. 

Ans,  I,  63.425  grams;  Na2S203,  5H2O,  124.15 
grams;  SO 2,  16.015  grams. 

12.  What  is  the  strength  of  a  ^/jo  sodium  thiosulphate 
solution  in  terms  of  iodine  ?  of  copper  ? 

Ans,     0.012685.     0.00636. 

13.  What  is  the  strength  of  a  ^/^  iodine  solution  in 
terms  of  SO2?  H2SO3  ?  Na2S03  ? 

Ans.     0.006406.     0.008208.     0.012616. 

14.  How  many  c.c.  of  ^Vs  Na2S203  solution  will  be 
required  to  react  with  the  iodine  liberated  by  20  c.c.  of  a 
^/lo  KaCrgOy  solution?  Ans,      10  c.c. 

15.  How  many  grams  of  copper  will  give  when  pre- 
cipitated by  potassium  iodide  sufficient  iodine  to  require 
20  c.c.  of  N/10  SO2  solution?  How  many  for  10  c.c.  of 
^/lo  Na2S203  solution?         Ans.     0.1272  and  0.0636. 


EXAMPLES.  89 

1 6.  Five  grams  of  pig  iron  took  12  c.c.  of  ^/loo  iodine 
solution.     What  is  the  percentage  of  sulphur  ? 

Ans.     0.0384%. 

1 7.  In  one  gram  of  an  antimony  alloy,  the  antimony 
was  determined  by  ^/jo  Na2S203,  24.2  c.c.  were  required. 
What  is  the  percentage  of  antimony?  Ans.     14.54%. 

18.  What  amount  of  type  metal  must  be  taken  for 
analysis  so  that  the  burette  shall  read  percentage  of 
antimony  when  ^/lo  NagSgOg  is  used? 

Ans.     0.601  gram. 

19.  Five  grams  of  bleaching  powder  were  mixed  in 
a  mortar  with  water  and  diluted  to  one  liter :  50  c.c.  of 
this  required  30  c.c.  of  ^/jq  NasAsOg  solution.  What  is 
the  percentage  of  available  chlorine  ?     The  reaction  is 

CaClsO  +  NasAsOg  =  CaCla  +  NagAsO^. 

Ans.     42.54%- 

20.  Under  the  conditions  of  the  preceding  example 
what  fraction  of  normal  will  read  percentage  direct  ? 

Ans.     ^ . 

14.18 

21.  To  50  c.c.  of  a  solution  of  chlorine  an  excess  of 
potassium  iodide  was  added;  the  liberated  iodine  was 
then  estimated  by  a  ^ I iq  solution  of  NagSgOs  using  starch 
as  an  indicator ;  22.5  c.c.  of  NaoSgOg  solution  were  used. 
What  is  the  strenorth  of  the  chlorine  solution  ? 

Ans.     0.001595  gram  per  c.c. 

22.  Five  grams  of  pig  iron  were  treated  with  hydro- 
chloric acid  and  the  HgS  evolved  passed  into  50  c.c.  of 
^7 10  NagAsOg  solution :  the  arsenious  sulphide  was 
filtered  and  washed,  and  the  filtrate  titrated  with  ^7 10 
iodine,  requiring  35  c.c.  What  was  the  percentage  of 
sulphur  in  the  sample  ?  Ans.     0.721%. 


90  EXAMPLES. 

23.  In  titrating  0.5  gram  of  an  iron  ore  by  KgCrgOy 
the  end  point  was  passed  when  60  c.c.  of  ^/lo  solution 
were  used.  One  gram  of  Mohr's  salt  was  added  and 
the  titration  continued  using  6.5  c.c.  more.  What  was 
the  percentage  of  iron  ?  Ans.     45.84%. 

24.  How  much  water  must  be  added  to  2  liters  of 
stannous  chloride  solution  containing  1 5  grams  of  tin  per 
liter  to  make  it  equivalent  to  a  K2Cr207  solution  which 
reads  percentage  of  iron  directly  when  0.5  of  a  gram  is 
taken  for  analysis  ?  Ans.      3637  c.c. 

25.  To  50  c.c.  of  a  stannous  chloride  solution  two 
grams  of  iron  were  added,  as  ferric  chloride,  the  solution 
was  then  titrated  by  ^/lo  KMn04  35  c.c.  being  required. 
How  much  SnClg  per  c.c.  did  the  solution  contain  ? 

Ans.     0.006646  gram. 


CHAPTER  VII. 

CALCULATIONS   OF    DENSITY   OF    SOLIDS   AND    LIQUIDS. 
DENSITY  OF  SOLIDS. 

The  density  of  solids  and  liquids  is  compared  with 
that  of  water  at  the  temperature  of  maximum  density. 
The  ratio  of  the  density  to  that  of  water  is  the  specific 
gravity.  So  if  we  say  a  substance  has  a  specific  gravity 
of  2.5,  we  mean  that  a  given  volume  has  2.5  times  the 
weight  of  the  same  volume  of  water  at  4°C  ;  or  that  its 
density  relative  to  that  of  water  is  as  2.5  is  to  i. 

The  density  of  solids  can  be  determined  by  the  fol- 
lowing methods.  The  usual  classification  is  given  here 
but  no  formulae,  as  these  are  more  confusing  than  help- 
ful and  do  not  lead  to  a  thorough  understanding  of  the 
subject. 

A.  Solids  heavier  than  water  and  insoluble. 

(i)  Weigh  in  air  and  then  in  water. 
According  to  the  law  of  Archimedes,  a  body  im- 
mersed in  a  fluid  loses  a  part  of  its  weight  equal  to  that 
of  the  fluid  it  displaces.  So  the  weight  of  the  solid  in 
air  minus  its  weight  in  water  gives  the  weight  of  an 
equal  volume  of  water;  as  the  volume  of  water  dis- 
placed is  equal  to  that  of  the  object,  unless  it  is  porous, 
when  it  must  be  allowed  to  soak  in  water  for  some  time  to 
expel  the  air.  So  if  we  divide  the  weight  in  air  by  the 
loss  in  weight  in  water,  we  have  the  specific  gravity  or 
density  compared  to  that  of  water  at  the  temperature  at 
which  the  experiment  was  carried  out.     This  can  then 

(91) 


52  DENSITY   OF   SOLIDS. 

if  necessary  be  corrected  for  the  expansion  of  water  and 
the  true  specific  gravity  compared  to  water  at  4°C  ob- 
tained. For  great  accuracy  the  weight  in  air  should  be 
reduced  to  weight  in  vacuo. 

For  example : —  Suppose  we  have  a  piece  of  brass 
whose  weight  in  air  is  24.076  grams  and  in  water 
21.2436  grams.  The  weight  in  air  divided  by  the  loss 
in  weight,  2.8324  grams,  gives  the  specific  gravity,  8.5, 
referred  to  that  of  water  at  the  temperature  of  the  ex- 
periment. 

(2)  If  the  substance  is  in  small  pieces  or  is  a  powder 
the  following  method  is  used : 

(a)  Weigh  the  substance  in  air. 

(b)  Weigh  a  small  flask  filled  either  entirely  with 
water  or  to  a  certain  mark. 

(c)  Place  the  substance  in  the  flask,  remove  air,  if 
necessary,  by  boiling  and  weigh  with  the  water  at  the 
same  point  as  before  (either  filled  or  to  the  mark). 

It  is  evident  that  a  volume  of  water  equal  to  that  of 
the  substance  has  been  displaced :  to  get  the  weight  of 
this  volume  of  water;  weight  of  substance  (a)  +  weight 
of  flask  and  water  (b)  —  weight  of  substance  in  flask 
and  water  (c)  =  weight  of  water  displaced.  Then  weight 
of  substance  in  air  divided  by  the  weight  of  an  equal 
volume  of  water  gives  specific  gravity.     For  example  : 

Weight  of  powder,  7  grams. 

Weight  of  flask  and  water  (to  mark)  55  grams. 

Weight  of  flask,  powder  and  water  (to  mark)  59 
grams. 

We  have   55  +   7  —  59  =  3  weight   of  volume   of 

water  displaced  by  powder:   and  -^=2.333,  the  specific 

3 
gravity  of  the  powder. 


DENSITY   OF   SOLIDS.  93 

By  having  graduations  on  the  neck  of  the  flask,  the 
volume  of  the  substance  may  be  read  off  by  the  rise  of 
the  water  in  the  flask,  then  this  volume  in  c.c.  multiplied 
by  the  weight  of  i  c.c.  of  water  at  the  temperature,  gives 
the  weight  of  an  equal  volume  of  water. 

B.  Solids  lighter  than  water  and  insoluble. 

As  these  solids  are  lighter,  they  cannot  be  weighed 
in  water  without  a  sinker. 

The  data  necessary  is — 

Weight  of  substance  in  air. 

Weight  of  sinker  in  water. 

Weight  of  substance  and  sinker  in  water. 

Or  if  we  have  the  weight  of  sinker  in  air  and  its 
specific  gravity  we  can  calculate  the  weight  of  the  sinker 
in  water. 

As  the  substance  is  lighter  than  water,  the  weight 
of  sinker  and  substance  in  water  is  less  than  the  sinker 
alone. 

If  we  subtract  from  weight  of  sinker  and  substance 
in  water  the  weight  of  sinker  in  water,  we  have  the 
weight  of  substance  in  water,  a  negative  quantity  and 
the  measure  of  its  buoyant  power. 

Now  the  weight  in  air  divided  by  the  weight  in  air 
minus  the  weight  in  water  is  the  specific  gravity,  but 
the  weight  in  air  minus  the  negative  weight  in  water 
equals  the  weight  in  air  plus  the  buoyant  power,  so  we 
have  in  this  case  the  specific  gravity  is  equal  to  the 
weight  in  air  divided  by  the  weight  in  air  plus  the  differ- 
ence in  weights  of  the  sinker  in  water  and  the  substance 
and  sinker  in  water. 

To  illustrate  by  an  example. — 

Weight  of  substance  in  air  4  grams. 

Weight  of  sinker  in  water  9  grams. 


94  DENSITY   OF   SOLIDS. 

Weight  of  substance  and  sinker  in  water  8  grams. 

Difference  in  weight  of  sinker  in  water  and  substance 
and  sinker  in  water  is  one  gram :  weight  in  air  plus  this 
difference  (or  minus  the  negative  loss  in  weight  of  one 
gram)  is  5. 

^.=  0.8  the  specific  gravity  of  the  substance. 

If  we  have  the  weight  of  the  sinker  in  air  and  its 
specific  gravity. 

The  weight  in  air  divided  by  the  specific  gravity 
gives  the  volume,  or  expressed  in  grams,  the  weight  of 
an  equal  volume  of  water.  This  is  equal  to  the  loss  of 
weight  in  water. 

The  weight  in  air  minus  the  loss  of  weight  in  water 
gives  the  weight  in  water. 

For  example,  suppose  the  sinker  weighed   1 2  grams 

and  its  specific  gravity  was  8. 

12.. 

— -=  1.5  (loss  in  weight)  and  12  -  1.5  or  10.5  =  weight  of 
8 

sinker  in  water. 

C    Solids  heavier  than  water  but  soluble. 

Weigh  in  air  then  in  a  liquid  of  known  specific 
gravity  in  which  the  substance  is  insoluble,  such  as  tur- 
pentine, petroleum,  alcohol,  &c. 

The  weight  in  air  divided  by  the  loss  in  weight  gives 
the  density  relative  to  the  liquid  used  ;  suppose  this  is 
7  and  the  specific  gravity  of  the  liquid  0.8,  to  find  the 
specific  gravity  we  have 

7 :  X  : :  I  :  0.8  or  7  x  0.8  =  5.  6  Sp.  Gr. 
Or,  the  loss  in  weight  of  the  solid  equals  the  weight  of 
liquid  displaced.     Let  this  be  x  and  y  the  weight  of  the 
same    volume   of  water;    then  x  :  y  :  :  0.8  :  i   and  the 
weight  in  air  divided  by  y  gives  the  specific  gravity. 


DENSITY   OF   SOLIDS.  95 

To  determine  the  specific  gravity  of  rock  salt    Given : 
Weight  of  sample  in  air  10.436  grams 
Weight  of  sample  in  turpentine  6.3547  grams 
Specific  gravity  of  turpentine  0.86 
10.436  -  6.3547  =  4.0813  loss  in  weight  in  turpentine 
equal  to  the  weight  of  an  equal  volume  of  turpentine. 
Let  y  =  weight  of  an  equal  volume  of  water 
4.0813  :  y  :  :  0.86  :  i.    y  =  4.7457  and 

— '-^^  =  2.109  Sp.  Gr.  of  rock  salt. 

4.7457 

or  — '"^^    =2.5572  density  relative  to  turpentine  and 

2.5572  :  x: :  I  :  0.86.     x  =  2.199  Sp.  Gr.  of  rock  salt 

D.  Solid  lighter  than  water  and  soluble. 

Weigh  with  a  sinker  in  some  liquid  which  does  not 
act  on  the  substance  and  calculate  the  density  relative 
to  the  liquid;  then  from  the  specific  gravity  of  the 
liquid  calculate  the  specific  gravity  as  under  C. 

To  determine  the  specific  gravity  of  lithium — 

Weight  of  lithium  in  air  3.  grams 

Weight  of  sinker  in  air    10.  grams 

Weight  of  lithium  and  sinker  in  kerosene  7.9222 
grams. 

Specific  gravity  of  sinker  (silver)  10.5 

Specific  gravity  of  kerosene  used  0.829 

First  find  the  specific  gravity  of  the  sinker  referred 
to  kerosene 

0.829  :  I  :  :  10.50  :  x.     x  =  12.6658 

Next  find  the  weight  of  the  sinker  in  kerosene    we 

have  — ^^  =  12.6658.      y  =  9.2104 

10-  y 


96  DENSITY   OF   LIQUIDS. 

Now  followingf  the  method  for  solids  lighter  than 
water  and  insoluble 

^ =  0.6996  density  of  lithium 

3  +  9-2104-7.9222 

relative  to  kerosene :  and  to  refer  to  water  we  have, 

0.829  :  I  :  :  z  :  0.6996.      z  =0.57996 

The  specific  gravity  of  lithium  is  therefore  0.58. 

In  these  examples  no  particular  attention  has  been 
given  to  the  temperature  of  the  water,  but  the  specific 
gravity  has  simply  been  referred  to  water  at  the  same 
temperature  (that  of  the  experiment).  To  refer  these 
specific  gravities  to  water  at  4°C;  look  up  in  the  table, 
at  the  back  of  the  book,  the  weight  of  a  cubic  centimeter 
of  water  at  the  particular  temperature  used,  then  substi- 
tute in  the  proportion, 
Sp.  Gr.x°C  :  Sp.  Gr.  4°C  :  :  i  :  wt.  i  c.c.  water  at  x°C.  * 

This  is  the  same  proportion  used  for  the  conversion 
of  the  relative  density  of  a  solid  in  turpentine  to  density 
referred  to  water  (specific  gravity)  and  it  is  a  similar 
proposition;  for  water  at  say  20°C  is  a  liquid  lighter 
than  water  at  4°C  so  the  specific  gravity  referred  to 
water  at  4°C  will  necessarily  be  less. 

The  correction  for  the  weight  of  the  solid  in  air  is 
too  small  to  have  significance  here  and  will  be  discussed 
in  the  next  chapter. 

DENSITY    OF    LIQUIDS. 

The  specific  gravity  of  liquids  is  determined  as 
follows : 

1st.   By  the  pyknometer  or  specific  gravity  bottle. 

*  The  table  also  gives  the  volume  of  one  gram  of  water  at  different  tempera- 
tures. To  use  these  values  substitute  in  the  proportion  Sp.  Gr.  X°C  :  Sp.  Gr. 
4°C  :  :  vol.  at  X°C  :  1. 


DENSITY   OF   LIQUIDS.  97 

2d.  By  weighing  a  solid  of  known  specific  gravity  in 
the  h'quid. 

3d.  By  hydrometers. 

4th.  By  Westphal's  balance.  This  is  really  only  a 
modification  of  the  second  method. 

I  St.  In  the  pyknometer  we  can  compare  directly  the 
vv^eights  of  the  same  volume  of  the  liquid  and  of  water  ; 
so  that  the  calculation  only  involves  the  subtraction  of 
the  weight  of  the  pyknometer  and  the  division  of  the 
weight  of  the  liquid  by  that  of  an  equal  volume  o/  water. 

2d  If  we  have,  for  example,  a  piece  of  brass,  which 
weighs  in  air  lo  grams,  in  water  8.824  grams  and  in  the 
liquid  of  unknown  specific  gravity  9  grams :  we  have  in 
each  case  the  loss  of  weight  of  the  brass  equals  the 
weight  of  the  volume  of  liquid  displaced  ;  and,  as  the 
volume  is  the  same  in  both  cases,  the  loss  in  weight  is 
directly  proportional  to  the  specific  gravity  and  we  have 

1. 1 76        :  I  :  :  i         :  x.       x  =  0.85 

Loss  of  wt.     Loss  of  wt.     Sp.  Gr.  of  Sp.  Gr.  of 
in  water.        in  Liquid.  water.  liquid. 

Or  if  we  know  the  weight  of  the  brass  in  air  and  its 
specific  gravity  we  can  calculate  the  loss  of  weight  in 
water  as  already  explained  under  the  density  of  solids. 

3d  Hydrometers  are  made  either  to  read  specific 
gravity  direct,  like  the  lactometer^  or  are  graduated  ac- 
cording to  some  arbitrary  scale  of  degrees. 

Degrees  Beaume  and  specific  gravity  may  be  com- 
pared by  reference  to  the  tables  at  the  end  of  the  book. 

Degrees  Twaddell  have  a  direct  relation  to  specific 

gravity.      200°  represent  the  densities  between  i    and  2 

so  I  °  represents  a  difference  of  specific  gravity  of  0.005, 

'c  V    •        1    N°  (Twaddell). 

hence  the  specific  gravity  is  i  + ^^ ^ 

^  200 


98  DENSITY   OF   LIQUIDS. 

36°  Twaddell  is  1.18  Sp.  Gr. 

94°  Twaddell  is  i  .47  Sp.  Gr.  and  so  on. 

4th.  In  using  the  Westphal's  balance  each  weight 
gives  the  figure  for  a  decimal  place  beginning  with  the 
largest.  If  the  liquid  is  heavier  than  water  a  weight 
(largest  size)  must  be  placed  on  the  hook  to  help  sink 
the  float.  This  weight  gives  i  (before  the  decimal  point) 
and  the  remaining  weights  the  decimals. 

Under  the  specific  gravity  of  liquids  come  the  prob- 
lems of^dilution  to  a  certain  specific  gravity.  This  can 
of  course  be  done  without  any  calculation  by  the  use  of 
a  hydrometer.  The  following  method  will  however  be 
found  useful,  although  not  strictly  accurate  on  account 
of  a  small  contraction  of  volume  on  mixing  which  is 
found  by  very  accurate  measurements.  * 

Suppose  we  wish  to  make  200  c.c.  of  ammonia  water 
0.96  Sp.  Gr.  and  we  have,  ammonia  water  0.88  Sp.  Gr. 
and  water.  The  ratio  is  inver,aely:.as  the  difference  in 
specific  gravities  .08  (.96  —  .88)®:  .04  (i.oo  —  .96)  :  : 
volume  of  water  :  volume  of  ammonia  r  two  parts  of 
water  to  one  of  strong  ammonia,  or  66.66  c.c.  diluted 
to  200. 

Required  to  make  six  liters  of  nitric  acid  1.26  Sp.  Gr. 
having  nitric  acid  i  .42  Sp.  Gr.  and  water. 

26  :  16  :  :  volume  of  acid  :  volume  of  water  :  or  16 
parts  of  water  for  42  of  diluted  acid  :  for  6  liters,  16  :  42 
:  :  X  :  6.  x  =  2.285.  ^^»  ^^^  Y  =  volume  of  nitric  acid 
in  liters  1.42  y  +  6  — y  =  6  (1.26).     y  =  3.714. 

These  methods  can  be  applied  to  any  case  of  mixing, 
as  the  contraction  previously  mentioned  will  not  intro- 
duce any  error  appreciable  in  ordinary  work. 

*  See  Wade,  Journal  Chemical  Society  p.  254,  1899. 


DENSITY   OF  LIQUIDS.  99  3 

Suppose  we  wish  to  make  up  two  liters  of  a  stan-  '4,  ^ 

dard  sulphuric  acid  solution  containing  ten  grams  of         V  I 

H0SO4  per  liter.     Dilute  pure  concentrated  sulphuric      >  /^  \ 

acid  with  water  to  about   1.7  Sp.  Gr.  allow  to  cool  to       L  V  A  ' 

I5°C  and  then  take  the  specific  gravity  most  accurately,  f  v>J    ?V  ! 

suppose  it  to  be  1.689.     The  percentage  of  H2SO4  can  s,  ^  ^^^  | 

be  found  by  referring  to  a  table*  to  be  76. 29.    The  weight  >  ^*  ^  i 

of  sulphuric    acid  to   be  diluted   will  be  -f^  />N  i 

76.29  J  ; 

=  26.2157  grams.    This  when  diluted  with  water  to  two    ^  i 

liters  at  15°  C  will  give  a  solution  containing  ten  grams  j 

of   H2SO4  per  liter.        In  this  calculation  the  specific  ' 

gravity  is  referred  to  water  at  15°  C  as  unity,  this  is  to  | 

agree  with  the  liter  used  in  dilution  (volume  of  1000  \ 

grams  of  water  at  15°  C).     Should  we  wish  to  compare  \ 

this  with  other  specific  gravities  it  can  be  reduced  to  " 

specific  gravity  at  4°  C  as  already  explained.  ; 

This  brings  us  to  the  consideration  of  changes  in  i 

specific  gravity  due  to  temperature.  { 

With   solids    the    change    in    volume   is   so   small  ] 

that  no  appreciable  error  is  introduced  by  neglecting^  ' 

this  variation   within    the  limits  of  laboratory  temper-  j 

ature.  | 

With  liquids  the  variations  of  specific  gravity  due  to  I 

changes  in  temperature  are  greater  and  cannot  be  dis-  ; 

regarded.     Unless  specially  stated  the  specific  gravity  ■ 
is  referred  to  water  ct  4°  C  as  unity  and  for  accurate 

work  the  determination  must  be  carried  out  at  that  tern-  ' 

perature.     For  many  important  solutions  and  liquids  the  ' 
corrections  for  temperature  have  been  worked  out  and 
are  to  be  found  in  the  books  on   technical  chemistry. 

*  Marshall,  J.  Society  Chemical  Industry  vol.  18,  p.  5.  1899. 


100  VOLUME   OCCUPIED   BY   PRECIPITATES. 

Thorpe^s  Dictionary  of  Applied  Chemistry  gives   the 
following  variations  of  sulphuric  acid 

o°C         io°C         20°  C         30°  C         40°  C         50°  C 
1.857        1-846  1.835  1-825  1. 816  1.806 

In  many  cases  the  correction  is  given  for  1°  C  over 
a  range  of  temperatures :  these  are  added  to  the  specific 
gravity  at  the  temperature  of  the  experiment  to  obtain 
the  gravity  at  4°  C* 

VOLUME    OCCUPIED    BY    PRECIPITATES. 

In  analyses  such  as  that  of  pig  lead  for  bismuth, 
the  question  of  the  volume  occupied  by  a  precipitate  be- 
comes of  importance. 

Suppose  seventy-five  grams  of  impure  lead  is  dis- 
solved in  nitric  acid,  the  lead  precipitated  by  sulphuric 
acid,  diluted  to  one  liter  and  thoroughly  mixed.  To 
save  time  in  filtering  and  washing,  it  is  desired  to  draw 
off  for  the  determination  of  bismuth,  such  a  volume  of 
the  solution  as  shall  represent  fifty  grams  of  the  original 
sample. 

Two-thirds  of  (1000  c.c.  minus  the  volume  of  the  pre- 
cipitate) will  be  the  volume  required.  The  amount  of  im- 
purities is  small  and  the  solubility  of  lead  sulphate  will 
not  introduce  an  appreciable  error,  so  we  assume  that 
the  weight  of  lead  sulphate  will  be  proportional  to  the 
seventy- five  grams  of  pig  lead. 

75  :  X  : :  206.9  •  302.96.     x  =  109.82  grams. 
The  specific  gravity  of  PbS04  is  6.2  therefore  the  volume 

.    ,  ,       ,  .  .         .    109.82  ^ 

occupied  by  the  precipitate  is  —^ —  or  17.7  c.c.     bo  to 

*  For  a  table  of  corrections  for  alcohol  see  Zeit.  fiir  Analytische  Chemie,  vol.  38, 
p.  253,  1899;  and  Morley,  J.  Am.  Chem,  See,  vol.  26,  p.  1 185,  1904. 


VOLUME   OCCUPIED    BY  PRECIPITATES.  101 

get  a  solution  containing  the  bismuth  from  fifty  grams  of 
pig  lead,  we  must  draw  off  or  filter  through  a  dry  filter 
73  (looo  —  17.7)  or  654.86  c.c. 

If  we  have  a  solution  containing  a  precipitate  and 
wish  to  titrate  the  excess  of  precipitant  in  a  portion  of 
the  solution,  we  dilute  to  the  mark  in  a  flask,  mix,  and 
then  draw  off  a  portion  with  a  pipette.  Suppose  the 
solution  is  diluted  to  200  c.c.  and  50  c.c.  withdrawn; 
this  is  evidently  more  than  one  quarter  on  account  of 
the  volume  occupied  by  the  precipitate.  This  error  is 
usually  disregarded,  but  it  can  be  found  as  follows.  If  we 
have  precipitated  the  lead  from  one  gram  of  lead  ore  as 
PbCr04,  by  a  known  amount  of  KoCro07,  and  wish  to 
determine  the  excess  by  titration  with  potassium  iodide 
and  sodium  thiosulphate  in  a  portion  of  the  filtrate ;  in- 
stead of  filtering,  washing  and  then  diluting  to  a  known 
volume  we  may  proceed  as  follows :  after  adding 
K2Cr207  in  excess,  dilute  to  200  c.c.  mix  thoroughly  and 
allow  the  precipitate  to  settle;  then  withdraw  50  c.c.  of 
the  clear  solution  and  titrate  the  excess  with  sodium  thio- 
sulphate after  adding  potassium  iodide  and  hydrochloric 
acid.  This  shows,  we  will  assume,  an  excess  of  1 00  mgs.  of 
K2Cr207 ;  on  the  assumption  that  we  have  titrated  one 
quarter  of  the  solution,  the  total  excess  of  KoCroOj 
will  be  400  mgs.  This  is  subtracted  from  the  known 
weight  of  KgCroOy  added  and  the  lead  present  calcu- 
lated from  the  remainder  in  the  usual  way,  giving  say 
80%  of  lead. 

Now  that  we  have  determined  approximately  the 
lead  present,  we  can  from  this  calculate  the  volume  occu- 
pied by  the  lead  chromate  and  so  correct  this  percentage 
of  lead  for  the  volume  occupied  by  the  precipitate.  80% 
Pb  =  0.8  gram  Pb  in  one  gram  =  1.249  grams  PbCrOi- 


102  VOLUME   OCCUPIED   BY   PRECIPITATES. 

The  specific  gravity  of  PbCr04  is  6,  hence  the  volume 

.    ,   ,       ,  .   .         .    1.240 

occupied  by  the  precipitate  is  — ^-  or  0.21  c.c.  50  c.c: 

showed  an  excess  of  100  mgs.  of  KaCrgOy,  to  get  the  cor- 
rect total  excess,  we  have —  50  :  200  —  0.21  :  :  100  :  x. 
X  =  399-58  mgs.  or  0.42  milligram  more  KgCrgOy  was 
used  in  precipitating  PbCr04  than  was  found  before. 
Which  is  equivalent  to  0.59  milligram  of  lead,  so  that 
the  correct  weight  of  lead  present  was  0.80059  gram, 
equivalent  to  80.059%. 

It  will  be  observed  that  this  correction  depends 
mainly  on  two  things — the  excess  of  reagent  and  the 
dilution.  If  the  excess  is  small  and  the  dilution  consid- 
erable, the  error  due  to  the  volume  of  the  precipitate  can 
be  neglected. 

The  accuracy  of  this  method  of  calculating  the 
volume  of  the  precipitate  is  questioned  by  Ostwald,  on 
account  of  the  variation  of  the  specific  gravity  of  the 
precipitate  in  the  solution,  due  mainly  to  adsorption. 
The  method  however  is  much  used  in  commercial  analy- 
tical work  with  satisfactory  results. 


EXAMPLES. 


1.  A  piece  of  barite  (heavy  spar)  weighs  in  air  47 
grams  and  in  water  36.555  grams.  What  is  its  specific 
gravity?  Ans,     /^.$. 

2.  A  specimen  of  magnetite  weighed  in  air  27  grams 
and  in  water  21.60  grams.    What  is  its  specific  gravity? 

Ans.     5. 

3.  A  piece  of  loaf  sugar  weighed  in  air  7.5  grams 
and  in  petroleum  ether,  Sp.  Gr.  0.645,  4.4635  grams. 
What  is  its  specific  gravity  ?  Ans.     1.593. 

4.  A  sample  of  magnetic  sand  weighed  in  air  12 
grams.  A  flask  filled  with  water  to  a  mark  weighed 
47.5327  grams.  The  same  flask  containing  the  sand 
and  filled  to  the  same  mark  with  water  weighed  56.4281 
grams.     What  is  the  specific  gravity  of  the  sand  ? 

Ans.     3.864. 

5.  To  find  the  specific  gravity  of  a  piece  of  freshly 
cut  pine  wood.     Given  : 

Weight  in  air,  ....  10.100  grams 

Weight  of  sinker  in  water,  .         .  12.24  grams 

Weight  of  wood  and  sinker  in  water,  1 1.2 17  grams 

Ans.     0.908. 

6.  To  find  the  specific  gravity  of  common  salt,  given 
the  following  data : 

Weight  of  salt  in  air,  .         .         .         8.8450  grams 

Weight  of  flask  and  turpentine  to  mark,   45-2375  grams 

(103) 


104  EXAMPLES. 

Wt.  of  flask,  salt  and  turpentine  to  mark,  50.5609  grams 
Specific  gravity  of  turpentine,  0.86 

Ans.     2.16 

7.  To  find  the  specific  gravity  of  metallic  sodium. 
Given  : 

Weight  in  dry  air,  .  .  .  14.15629  grams 

Weight  in  kerosene,  .  .         .  2.1807  grams 

Specific  gravity  of  kerosene,  .  .         .         0.829 

Ans.     0.98. 

8.  To  find  the  specific  gravity  of  turpentine.  Given: 
Weight  of  pyknometer,  .  .  .  25.0456  grams 
Wt.  of  pyknometer  filled  with  water,  .  72.1485  grams 
Wt.  of  pyknometer  filled  with  turpentine,  65.5541  grams 

Ans.      0.86. 

9.  A  piece  of  silver  weighed  47  grams  in  water  and 
48.756  grams  in  petroleum  ether,  specific  gravity  0.645. 
What  is  the  specific  gravity  of  the  silver? 

Ans.      10.5. 

10.  A  ten  gram  brass  weight,  Sp.  Gr.  8.5,  weighed 
in  glycerin  8.5135  grams.  What  is  the  specific  gravity 
of  glycerin?  Ans.      1.2635. 

11.  Given  the  specific  gravity  of  sulphuric  acid  as 
1.84  compared  to  water  at  15°  C.  What  is  its  specific 
gravity  compared  to  water  at  4°  C  ?     (See  tables.) 

Ans.     1.8384. 

.  1 2.  Given  the  specific  gravity  of  alcohol  at  20°  C 
referred  to  water  at  4°  C  as  0.8332.  What  is  its  specific 
gravity  referred  to  water  at  20°  C  ?      Ans.     0.83467. 

13.  How  much  phosphoric  acid,  1.7  Sp.  Gr.  and 
how  much  water  are  required  to  make  400  c.c.  of  1.18 
Sp.  Gr.  ? 

Ans.  102.85  c.c.  phosphoric  acid  and  297.15  c.c.  of 
water. 


EXAMPLES.  105 

14.  How  much  nitric  acid  1.47  Sp.  Gr.  and  water 
must  be  mixed  to  give  three  liters  of  1.08  Sp.  Gr.  ? 

Ans.     510.6  c.c.  nitric  acid;     2489.4  c.c.  water. 

15.  How  much  ammonia  water  0.9  Sp.  Gr.  must  be 
taken  to  make  with  water  four  liters  0.96  Sp.  Gr.  ? 

A71S,     1.6  liters. 

16.  We  have  two  liters  of  dilute  sulphuric  acid  1.08 
Sp.  Gr.,  which  we  wish  to  use  up  in  diluting  concen- 
trated sulphuric  acid  1.82  to  1.20  Sp.  Gr.  How  much 
concentrated  acid  is  to  be  added?       Ans.     387.1  c.c. 

1 7  What  is  the  volume  occupied  by  47  grams  of 
PbS04.     Sp.  Gr.  6.2  ?  Ans.     7.5  c.c. 

18.  How  many  c.c.  of  sulphuric  acid,  1.824  Sp.  Gr. 
compared  to  water  at  15°  C,  weigh  100  grams? 

Ans.     54.88  c.c. 

19.  What  is  the  volume  occupied  by  the  barium  sul- 
phate from  one  gram  of  FeSo?     Sp.  Gr.  BaS04  4.5. 

Ans.     0.86  c.c. 

20.  To  two  grams  of  a  mineral  containing  barium, 
200  mgs.  of  KgCroOy  were  added  to  precipitate  the 
barium  and  the  solution  diluted  to  500  c.c,  mixed  and 
allowed  to  settle  :  50  c.c.  of  the  clear  solution  was  titrated 
for  the  excess  of  KgCroOy  by  KI  and  NaoSoOg  showing 
4.2  mgs.  What  is  the  corrected  percentage  of  barium 
oxide?     Given  the  Sp.  Gr.  of  BaCr04  as  3.9. 

Ans.  8.23%.  Showing  the  correction  to  be  insig- 
nificant. 


CHAPTER   VIII. 

CALCULATIONS   OF   GASES. 

Before  proceeding  to  any  of  the  calculations  of  gases 
or  gas  analysis  a  knowledge  of  the  following  laws  is 
necessary. 

Law  of  Boyle  or  Mariotte. 

1.  The  temperature  remaining  the  same,  the  volume 
of  a  given  quantity  of  gas  is  inversely  as  the  pressure  it 
bears. 

Law  of  Charles  or  Gay-Lussac. 

2.  The  coefficient  of  expansion  of  all  gases  is  0.00366 

or  of  the  volume  at  0°  C. 

273 

Differently  expressed,  the  volume  of  all  gases  is  di- 
rectly proportional  to  the  temperature  calculated  from 
the  absolute  zero  (-273°  C). 

Law  of  Avogadro. 

3.  Under  the  same  temperature  and  pressure  equal 
volumes  of  all  gases  contain  the  same  number  of  mole- 
cules. 

It  is  evident  from  (i)  and  (2)  that  the  volume  of 
gases  varies  greatly  under  different  conditions  of  tem- 
perature and  pressure  so  that  these  must  always  be  re- 
duced to  the  standard  conditions  of  0°  C  and  760  milli- 
meters pressure  for  comparison. 

(106) 


CALCULATIONS   OF   GASES.  107 

This  is  done  as  follows :  Suppose  we  have  six  liters 
of  a  gas  at  i8°  C.  and  772  m.m.  pressure  and  we  wish 
to  know  the  volume  under  the  standard  conditions.  We 
can  first  correct  for  temperature  and  then  for  pressure  or 
vice  versa.  The  volume  varies  as  the  temperature  calcu- 
lated from  the  absolute  zero,  therefore 

273  •  273  +  iS  •  •  X  :  6.     X  =  5.6288. 
The  volume  at  0°  C  and  772  m.m.  is  5.6288  liters.    The 
volume  varies  inversely  as  the  pressure,  therefore : 
772  :  760  :  :  y  :  5.6288.     y  =  5.7176  liters, 
the  volume  of  the  gas  at  0°  C  and  760  m.m.  pressure. 

These  calculations  are  necessary  in  gas  analysis  and 
in  many  other  analytical  determinations  where  the 
volume  of  a  gas  is  measured.  When  the  gas  is  measured 
over  water  it  contains  aqueous  vapor  which  supports 
part  of  the  pressure:  this  tension  increases  with  the 
temperature  (see  table),  therefore  before  proceeding  to 
the  corrections  for  temperature  and  pressure  as  before, 
the  tension  of  aqueous  vapor  must  be  subtracted 
from  the  barometer  reading  to  get  the  real  pressure 
supported  by  the  gas.  Take  the  following  example  of 
a  nitrogen  combustion.  Weight  substance  o.  1923  gram  : 
volume  nitrogen  24.6  c.c. :  temperature  25°  C.  :  baro- 
metric pressure  771  m.m.  The  tension  of  aqueous  vapor 
at  25°  C  is  23.5  m.m.;  so  771  —  23.5  or  747.5  m.m.  is  the 
pressure  supported  by  the  nitrogen.  As  the  volume 
varies  inversely  as  the  pressure,  we  have : 

24.6  :  X  :  :  760  :  747.5.     x  =  23.93  c.c. 

(volume  at  normal  pressure) 
As  volume  varies  directly  as  the  temperature  calculated 
from  the  absolute  zero. 

23.93  :  y  :  :  273  +  25  :  273.    y  =  21.95  c.c. 


108  DENSITY   OF   GASES. 

volume  of  nitrogen  at  standard  conditions,      i  c.c.  nitro- 
gen at  760  m.m.  and  0°  C  weighs  0.001254  gram  (see 
table). 
Wt.  of  nitrogen   is  0.001254  X  21.95   =  0.02751   gram 

o  02  '1  ^ 
and  — — ^  X  100  =  14.31%  nitros^en. 
0.1923 

Suppose  we  have  200  c.c.  of  gas  at  10°  C  and  765 
m.m.  and  wish  to  know  the  volume  at  18°  C  and  772 
m.m. :  to  correct  for  temperature,  we  have : 

200  :  X  :  :  273  +  10  :  273  +  18.    x  =  202.08  c.c. 
and  for  pressure 

202.08  :  y  :  :  772  :  765.     y  =  200.24  c.c. 
the  volume  under  the  desired  conditions. 


DENSITY  OF  GASES. 

The  density  of  gases  is  referred  sometimes  to  air  at 
760  m.m.  pressure  and  0°  C.  as  unity  and  sometimes  to 
hydrogen  under  the  same  conditions  of  temperature  and 
pressure.  In  the  actual  determinations  it  is  usually  com- 
pared with  an  equal  volume  of  air  and  then  referred  to 
hydrogen  by  calculation.  Since  a  liter  of  hydrogen 
weighs  0.08973  gram  under  the  standard  conditions  and 
a  liter  of  air  1.29327  grams,  the  density  referred  to  air 
can  be  referred  to  hydrogen  by  multiplying  by  14.41. 

The  important  relation  between  the  molecular  weight 
and  density  of  gases  referred  to  hydrogen  has  already 
been  mentioned  under  the  calculation  of  atomic  weights, 
as  a  means  of  determining  what  multiple  of  the  equivalent 
is  the  atomic  weight ;  it  is  also  one  of  the  most  import- 
ant methods  for  determining  molecular  weight.     When 


DENSITY   OF  GASES.  109 

we  say  the  density  of  a  gas  is  40  referred  to  hydrogen, 
it  means  that  any  given  volume  of  the  gas  is  40  times  as 
heavy  as  the  same  volume  of  hydrogen,  under  the  same 
conditions  of  temperature  and  pressure.  Now  accord- 
ing to  the  law  of  Avogadro,  these  equal  volumes  of 
gases  contain  the  same  number  of  molecules  so  each 
molecule  of  the  gas  is  40  times  as  heavy  as  a  molecule 
of  hydrogen,  and  as  the  hydrogen  molecule  contains  two 
atoms,  a  molecule  of  the  gas  is  80  times  as  heavy  as  a 
hydrogen  atom  or  its  molecular  weight  is  80.  Or,  in 
brief,  the  density  of  a  gas  (referred  to  hydrogen)  is  one 
half  its  molecular  weight. 

The  exact  calculation  of  density  and  so  molecular 
weight  varies  with  the  method  used,  but  in  general  it  is 
as  follows : 

We  have  given  the  weight  of  a  known  volume  of 
gas  under  certain  conditions  of  temperature  and  pres- 
sure. Reduce  these  to  o°C  and  760  m.  m.,  and  divide 
the  weight  by  that  of  an  equal  volume  of  hydrogen 
under  the  same  conditions  and  we  have  the  density. 
The  density  of  vapors  is  determined  in  a  similar  way, 
except  that  the  volume  occupied  by  the  vapor  from 
a  known  weight  of  substance,  is  measured  instead  of 
the  weight  of  a  known  volume  of  vapor  or  gas.  In 
many  cases,  especially  when  the  heat  necessary  to 
convert  the  substance  into  the  gaseous  condition  is 
high,  it  is  more  convenient  to  measure  the  volume 
of  air  displaced  by  the  substance  in  the  gaseous  form, 
rather  than  the  gas  itself;  but  the  calculation  is  the 
same,  for  the  volume  occupied  by  the  air  is  the  same 
as  would  be  occupied  by  the  vapor  from  the  known 
weight  of  the  substance  at  the  same  temperature  and 
pressure. 


110  DENSITY   OF   GASES. 

The  gram-molecular  volume  is  the  volume  in  liters 
occupied  by  the  molecular  weight  in  grams  of  any  sub- 
stance in  the  gaseous  state  at  o°  C.  and  760  m.m.  pres- 
sure. Now  according  to  the  law  of  Avogadro  equal 
volumes  of  all  gases  contain  the  same  number  of  mole- 
cules or  conversely  an  equal  number  of  molecules  occupy 
the  same  volume,  so  that  the  gram-molecular  volume  of 
all  gases  is  the  same.  This  is  calculated  as  follows  : 
the  molecular  weight  of  oxygen  in  grams  is  32  :  the 
weight  of  a  liter  of  oxygen  is  1.429  grams  :  the  gram- 
molecular  volume  is  therefore  32  -f-  1.429  or  22.393  liters. 

So  we  may  consider  the  molecular  weight,  as  that 
weight  of  a  substance  in  grams,  which  in  the  state  of  a 
gas  at  0°  C.  and  760  m.m.  pressure  occupies  22.393 
liters,  or  in  round  numbers  22.4  liters  ;  or  dividing  by 
one  thousand,  the  weight  in  milligratms  which  occupies 
22.4  cubic  centimeters  under  these  conditions. 

This  conception  of  molecular  weight  is  not  only  im- 
portant in  itself  but  simplifies  the  calculation  of  the 
molecular  weight  determinations  where  volumes  of  gases 
are  measured,  as  will  be  seen  by  the  following  examples^ 
which  are  worked  out  by  both  methods. 

The  following  determinations  of  the  density  of  chloro- 
form vapor  by  the  three  principal  methods,  illustrate 
these  calculations. 

(i)  Dumas'  method.     Given  the  following  data. — 

Weight  of  bulb  filled  with  air  at  i9°C  and  763.9 
m.  m.  =  36.4489  grams 

Weight  of  bulb  filled  with  vapor  at  ioi°C  and  763.9 
m.  m.=  37.2085  grams. 

Volume  of  air  in  the  bulb  at  i9°C  =  279  c.c. 


DENSITY   OF  GASES.  Ill 

First  find  the  weight  of  the  air  contained  in  the 
bulb ;  by  reducing  the  volume  to  the  standard  conditions 
and  multiplying  by  0.001293,  the  weight  of  i  c.c.  of  air 
at  o°C  and  760  m.  m. 

279  :  X  :  :  273  +    19  :     273.     x  =  260.8  c.c. 
260.8  :  y  : :       760  :  763.9.     y  =  262.1  c.c. 

Weight  of  air  is  262.1  X  0.001293  =  0.3389  gram. 
Weight  of  bulb  is  36.4489  —  0.3389  =  36. 1 100  grams. 
Weight  of  vapor  is  37.2085  —  36.1100  =  1.0985  grams. 
We  now  know  that  279  c.c.  of  chloroform  vapor  at 
lOi^C  and  763.9  m.  m.  weighed  1.0985  grams. 

To  find  the  corresponding  volume  under  the  stand- 
ard conditions,  we  have  : 

279  :  X  : :  273  +  loi  :  273.     x  =  203.8  c.c. 
203.8  :  y  : :  760  :  763.9     y  =  204.8  c.c. 
The  weight  of  an  equal  volume  of  air  will  be 
204.8  X  0.001293  or  0.2648  gram. 

So 

1.0985  -i-  0.2648  or  4.148 

is  the  density  referred  to  air: 

4.148  X  14-41  or  59.773 
is  the  density  referred  to  hydrogen  and 
59-773  X  2  or  119.546 

is  the  molecular  weight.  Theory  119.36.  Or  1.0985 
grams  occupy  204.8  c.c.  The  molecular  weight  will 
occupy  22.4  liters.     Hence 

1.0985  :  X  : :  0.2048  :  22.4.     x  =  120.15. 

(2)  Victor  Meyer's  method.     Given  following  data. 


112  DENSITY   OF   GASES. 

Weight  of  chloroform  taken  0.2097  gram. 

Volume  of  air  expelled  by  vapor  43.65  c.c. 

Temperature  of  air  expelled  20° C. 

Height  of  barometer  763.9  m.  m. 

Tension  of  aqueous  vapor  at  20°C  (from  table)  17.36 
m.  m. 

We  have  the  volume  of  air  expelled  by  and  equal  to 
the  volume  of  the  vapor;  to  find  its  weight  reduce  this 
volume  to  the  standard  conditions  ; 

43.65  :  x  :  :  273  +  20  :  273.     x  =  40.67  c.c. 
40.67  :  y  :  :  760  :  763.9  -  17.36.      y  =  38.50  c.c. 

38.50  X  0.001293  =  0.04978  gram  :  weight  of  air. 

Then  0.2097  -f-  0.04978  is  4.21,  the  density  referred 
to  air ; 

4.21  X  14.41  =  60.67,  the  density  referred  to, 
hydrogen;  60.67  X  2  =  121.34,  the  molecular  weight. 
Or  as  0.2097  gram  occupied  38.5  c.c.  0.2097  :  x  :  : 
0.0385  :  22.4.     X  =  122.01. 

(3.)  Hofmann's  method.  The  following  data  are  re- 
quired : 

Weight  of  chloroform  o.  1 535  gram. 

Volume  of  vapor  at  ioo°C  79  c.c. 

Height  of  barometer  at  2i°C  774.8  m.  m. 

Height  of  column  of  mercury  (inside  heating 
jacket)  at  ioo°C  246.5  m.  m. 

Height  of  column  of  mercury  (outside  heating 
jacket)  at  25°C  143.8  m.  m. 

*  Tension  of  mercury  vapor  at  ioo°C  0.27  m.  m. 

This  calculation  involves,  besides  those  already  dis- 
cussed, the  reduction  of  the  barometric  readings  to  o°C  ; 
because  they  are  at  such  widely  varying  temperatures. 
The  coefficient  of  the  expansion  of  mercury  is  0.00018 

*  Ostwald  Physico-chemical  measurements  p.  106. 


DENSITY   BY   EFFUSION.  113 

for    i°C  :  so    to    reduce  the    reading-  representing   the 
atmospheric  pressure  to  zero,  we  have  : 

I  :  I  +  (21  X  0.00018)  :  :  X  :  774.8  or 
774.8  -i-  1.00378  =  771.88  m.  m. 

For  the  other  columns  of  mercury  we  have  : 

246.5  -i-  1. 018  =  242.14  m.   m.   and   143.8  -^  1-0045  = 

143.15  m.  m. 
The  79  c.c.  of  vapor  is  under  the  pressure  indicated 
by  the  barometer  minus  the  tension  of  mercury  vapor 
(0.27  m.  m.)  and  minus  the  sum  of  the  two  columns  of 
mercury  also  supported  by  the  atmospheric  pressure 
(242.14+  HS-^S)-  So  the  pressure  is  771.88  —  (0.27 
+242.14+143.15)  =  385.56  m.  m. 

We  now  know  that  0.1535  gram  of  vapor  at  ioo°C 
and  385.56  m.  m.  occupy  79  c.c.  Reduce  this  as  before 
to  o°C  and  760  m.  m  pressure  and  divide  the  weight 
by  that  of  an  equal  volume  of  air  : 

79  :  X  :  :  373  :  273.     x  =  57.82  c  c. 
57.82  :  y  :  :  760:  3^5-5^'     Y  =  29.333  c.c. 
.    29.333  X  0.001293  =  0.037927 
0.1535  -i-  0.037927  =  4.047  the  density  re- 
ferred to  air. 

DENSITY  BY    EFFUSION. 

The  density  or  specific  gravity  of  gases  can  be  deter- 
mined with  sufficient  accuracy  for  many  practical  pur- 
poses by  the  effusion  test.* 

This  is  based  on  the  fact  that  the  specific  gravity  of 
gases  is  proportional  to  the  squares  of  the  times  of  effu- 
sion.     In  these  tests  it  is  most  convenient  to  determine 


*  For  details  see  Hempel's  Gas  Analysis,  Dennis  p.  212. 


114  CORRECTION   OF   WEIGHINGS 

the  specific  gravity  with  reference  to  air  and  then,  if 
necessary,  refer  it  to  hydrogen  by  calculation. 

Suppose  a  certain  volume  of  air  requires  twenty 
minutes  to  escape  through  an  opening  made  by  a  needle 
in  a  platinum  foil,  and  that  under  the  same  conditions  of 
temperature  and  pressure  the  same  volume  of  another 
gas  requires  twelve  minutes. 

Let  X  =  density  of  this  gas  :  then 

X  :  I  :  :  144  :  400.  x  =  0.36,  density  referred  to 
air,  and  0.36  X  i4-4i  =  5- 188,  density  referred  to 
hydrogen. 

This  test  is  much  used  for  illuminating  gas  at  the 
works,  but  is  not  sufficiently  accurate  for  scientific 
work. 

CORRECTION    OF  WEIGHINGS. 

According  to  the  law  of  Archimedes,  when  any 
object  is  weighed  under  the  ordinary  conditions  in  air, 
we  obtain  less  than  the  true  weight  on  account  of  the 
loss  in  weight  of  the  object  equal  to  that  of  the  air  dis- 
placed. This  correction  is  small,  but  it  must  be  made 
in  some  cases,  especially  when  large  light  apparatus  is 
weighed,  such  as  absorption  bulbs  &c.,  and  in  all  very 
accurate  determinations,  such  as  atomic  weights. 

The  correction  is  therefore  the  addition  of  the  weight 
of  an  equal  volume  of  air  at  the  temperature  of  the 
weighing.  This  may  be  done  as  follows :  weight  in  air 
then  in  water ;  the  loss  in  weight  is  equal  to  the  weight 
of  the  same  volume  of  water ;  now,  at  the  ordinary 
laboratory  temperature  the  density  of  air  referred  to 
water  as  unity  is  0.0012,  so  the  loss  in  weight  of  the 
substance  in  water  times  0.0012*  is  the  correction  to  add. 

*  This  weight  of  1  c.c.  allows  for  mean  humidity  as  well  as  temperature. 


CORRECTION   OF   WEIGHINGS.  115 

If  the  specific  gravity  of  the  object  is  known  the  cor- 
rection can  be  calculated  as  follows : 

Wt.  in  air  *  , 

— T^ — -^ =  volume  in  c.c. 

bp.  Gr. 

Volume  times  weight  of  i  c.c.  of  air  (0.0012)  is  the 
correction  to  add. 

In  making  corrections  for  weighing  it  must  be  borne 
in  mind  that  we  have  : 

(a)  The  apparent  weight,  that  actually  found. 

(b)  The  true  weight  in  vacuo,  obtained  by  calcu- 
lation. 

To  get  the  true  weight  of  any  object,  we  must  add  to 
its  apparent  weight  the  weight  of  an  equal  volume  of  air 
and  subtract  the  weight  of  a  volume  of  air  equal  to  that 
of  the  weights  used.  The  correction  for  the  weights 
diminishes  the  correction  for  substances  having  a  lower 
specific  gravity  than  that  of  the  weights;  when  the  den- 
sities are  the  same  the  corrections  equalize  each  other : 
when  the  density  of  the  object  is  greater  than  that  of 
the  weights  the  correction  becomes  negative  (is  to  be 
subtracted). 

Suppose  we  have  a  100  gram  brass  weight  correct 
in  air ;  it  will  weigh  slightly  more  in  vacuo.  To  find 
this  weight  we  have  : 

rr            .  ,                       .       •  L.    I    ftrue  wt.  X  0.0012) 
True  weight  =  apparent  weight  -|-  ^ ^ — ^ ^ 

or   i^  X-^.0012  =  0.0141  ;   so  that  the  true 

weight  in  vacuo  is  100.01 41  grams. 

Let  us  use  this  weight  to  weigh  different  substances 
and  calculate  their  true  weight. 

*  The  method  given  is  all  that  is  ever  required  but  is  not  strictly  accurate, 
for  the  true  weight  should  be  used  instead  of  the  weight  in  air. 


116  CORRECTION   OF   WEIGHINGS. 

I  St.   Sulphur,  specific  gravity  2.     The  correction  will 

be  volume  X  0.00 1 2   or  —  X  o.ooi  2  =  0.06  so  that  the 

2 

true  wei^rht  becomes 

100+  (0.06  — 0.0I4I)  =  100.0459. 

We  see  from  this  that  the  true  weight  is  that  of  the 
brass  weight  in  air  plus  the  weight  of  air  displaced  by  the 
object  less  that  displaced  by  the  weight  itself 

2nd.  If  we  balance  the  100  gram  brass  weight  with 
brass  of  the  same  specific  gravity,  we   shall   have   the 

correction  to  add  -  —  X  0.0012  =  0.014  and  that  to  de- 
duct 0.014,  so  that  the  true  weight  coincides  with  the 
apparent  weight. 

3rd.  If  we  balance  the  100  gram  brass  weight  with 
platinum,  specific  gravity  21.5,  we  shall  find  the  true 
weight  less  than  100  grams ;  for  the  correction  to  add  is 

TOO 

X  0.0012  =  0.0056;   while  that  to  subtract  is  as  be- 


21.5 

fore  0.014 :   hence  the  true  weight  becomes  100  +  0.0056 

-0.014  =  99.9916  grams. 

It  will  be  seen  from  the  preceding  that  the  correction 
is  different  for  brass  and  for  platinum  weights,  and  that 
the  correction  becomes  greater  as  the  specific  gravity 
become  less. 

Suppose  we  have  a  glass  vessel  whose  apparent 
weight  is  75  grams  when  weighed  with  brass  weights, 
and  whose  specific  gravity  is  2.5.  What  is  the  true 
weight  ? 

The  volume  is  1^  or  30.  c.c. ;  the  weight  of  an  equal 
volume  of  air  is  30  X  0.0012  =  0.036  gram:  the  correc- 
tion for  the  brass  weights  is  ^^  X     0.0012    =    0.0105 

^•5 


CALCULATIONS   OF   GAS   ANALYSIS. 


117 


gram  :   so  the  true  weight  becomes  75  +  (0.036-0.0105) 
=  75-0255  grams. 

The  following  table  gives  the  weight  in  milligrams 
to  be  added  or  subtracted  for  each  gram  of  apparent 
weight  of  substance,  for  different  specific  gravities,  when 
weighed  in  air  with  brass  weights.* 


Sp.  Gr. 

Correction. 

Sp.  Gr. 

Correction. 

Sp.  Gr. 

Correction. 

0.7 

+  1.57 

2.0 

+   0.458 

9. 

—  0.009 

0.8 

+  1.36 

2.5 

+   0.337 

10. 

—  0.023 

0.9 

4-  1.19 

3.0 

+   0.257 

11. 

—  0.034 

1.0 

+  1.06 

3.5 

+   0.200 

12. 

—  0.043 

1.1 

+  0.95 

4.0 

+   0.157 

13. 

—  0.050 

1.2 

4-  0.86 

4.5 

+   0.142 

14. 

—  0.057 

1.3 

+  0.78 

5.0 

4-   0.097 

15. 

—   0.063 

1.4 

+  0.71 

5.5 

+   0.075 

16. 

—  0.068 

1.5 

+  0.66 

6.0 

+   0.057 

17. 

—  0.072 

1.6 

+  0.6L 

6.5 

+   0.042 

18. 

—  0.076 

1.7 

+  0.56 

7.0 

+   0.029 

19. 

—  0.080 

1.8 

+  0.52 

7.5 

+   0.017 

20. 

—   0.083 

1.9 

+  0.49 

8.0 

+   0.007 

CALCULATIONS    OF    GAS    ANALYSIS. 

With  the  help  of  the  law  of  Avogadro  we  can  very 
readily  understand  the  combination  of  gases  by  volume. 
If  we  cause  two  liters  of  carbonic  oxide  to  combine  with 
one  liter  of  oxygen,  the  product  is  not  three  liters  of 
carbonic  acid  but  two ;  for  there  are  present  the  same 
number  of  molecules  of  COo  as  there  were  of  CO  and  as. 
equal  volumes  of  all  gases  contain  the  same  number  of 
molecules,   an  equal   number  of  molecules  occupy  the 

*  Calculated  by  Kohlrausch.     Ostwald's  Physico-Chemical  Measurements. 


118  CALCULATIONS   OF   GAS   ANALYSIS. 

same  volume,  at  the  same  temperature  and  pressure : 
hence  we  have 

2  vols.  CO  +  I  vol.  Oo  =  2  vols.  CO2. 
This  shows  that  the  combination  of  the  two  gases  is  at- 
tended by  a  decrease  in  volume  and  that  the  resulting 
volume  is  two-thirds  of  that  occupied  by  the  mixture. 

Similarly  we  have 

2  vols.  H2  +  I  vol.  ©2=2  vols.  HgO  (as  steam) 

I  vol.  CH4  +  2  vols.  Oo  =  I  vol.  CO2+  2  vols.  H2O 
(as  steam)  or  if  the  steam  be  allowed  to  condense 

I  vol.  CH4  +  2  vols.  Oo  =  I  vol.  CO2  +  water. 

This  law  together  with  the  corrections  for  tempera- 
ture, pressure  and  tension  of  vapor,  already  explained, 
furnish  all  the  information  necessary  to  calculate  gas 
analyses. 

For  example  let  us  calculate  the  percentages  from 
the  following  analytical  results  on  a  sample  of  water 
gas. 

Volume  taken  for  analysis,         .  .         .  100  c.c. 

Contraction  due  to  absorption  of  CO2,         .  0.0  c.c. 

**  "    "  '*  ''  illuminants,       12.6  c.c. 

**  *'     ''  "  ''  oxygen,     .  0.9  c.c. 

*'  '*    **  "  *'  CO,  .        27.3  c.c. 

We  have  left  59.2  c.c.  of  the  original  100  c.c.  containing 

hydrogen,  methane  &c.,  and  nitrogen,    16.2  c.c.  of  this 

are  withdrawn  to  the  explosion  tube  and  mixed  with 

32.3  c.c.  of  oxygen  and  then  diluted  with  31.4  c.c.  of  air 
giving  a  total  volume  of  80.1  c.c. :  this  is  then  exploded 
and  after  allowing  to  cool  the  volume  is  found  to  be 

53.4  c.c. 

We  have  in  the  tube  before  explosion. 
X  Vols.  CH4  -f-  y  vols.  Hg  +  m  vols.  N2  +  n  vols.  Og. 


CALCULATIONS   OF   GAS   ANAYLSIS.  119 

After  explosion  this  becomes 

X  vols.   COo  +  (2  X  +  y)  vols.  HoO  (condensed)  +  m 

vols.  N2  (unaffected)  +  n  —  (2  x  +  J'^  y)  vols.  Og. 

The  volume  of  oxygen  is  diminished  by  a  volume 
equal  to  twice  that  of  the  CH4  plus  one-half  that  of  the 
hydrogen. 

The  CO2  produced  by  the  explosion  is  removed  by 
dilute  potassium  hydroxide  solution  and  the  volume  of 
the  gas  again  measured  =  45.8  c.c.  The  contraction  in 
volume  produced  by  the  removal  of  CO2  equals  the 
volume  of  CH4  present,  7.6  c.c.  The  contraction  in 
volume  due  to  the  explosion,  26,7  c.c,  therefore  repre- 
sents the  volume  occupied  by  the  hydrogen  (condensed 
to  water)  and  the  decrease  of  oxygen  equal  to  twice  the 
volume  of  CH4  plus  one  half  the  volume  of  hydrogen. 
The  CO2  produced  takes  the  place  of  the  CH4  originally 
present. 

If  we  let  y  =  vol.  of  hydrogen,  we  have : 
26.7  =  y  +  2(7.6)  +  %  y.     y  =  7.66  c.c. 
Or,  the  volume  of  hydrogen  =  ^  (of  contraction  minus 
twice  the  volume  of  CO2  formed)  (second  absorption  by 
KOH). 

As  only  a  portion  of  the  residual  gas  was  taken  for 
the  explosion,  the  volumes  obtained  are  referred  to  the 
original  100  c.c.  by  proportions: 

for  CH4,  16.2  :  59.2  :  :  7.6    :  X.     x  =  27.77  c.c. 
for  Ho,      16.2  :  59.2  :  :  7.66:  y.     y  =  27.99  c.c. 

The  results  expressed  in  percentage  by  volume  are : 
carbonic  acid,  none :  illuminants,  1 2.6%  :  oxygen,  0.9% : 
carbonic  oxide,  27.3%  :  hydrogen,  28%  :  methane  &c., 
27.7%:  nitrogen  (by  difference)  3-5%- 

For  accurate  work  all  volumes  should  be  corrected 


120  CALCULATIONS   OF   GAS   ANALYSIS. 

for  temperature,  pressure,  and  vapor  tension  as  already 
explained. 

The  following  are  results  on  coal  gas  with  the  Elliott 
apparatus : 

Volume  taken  for  analysis  loo  c.c. 

Reading  after  treatment  with  KOH.  (COo)  o.o  c.c. 
**  **  **  **      bromine  (illuminants) 

6.5  c.c. 

Reading  after  treatment  with  alkaline  pyrogallate 
(O2)  7.4  c.c. 

Reading  after  treatment  with  CuoClg  and  HCl 
(CO)  14.2  c.c. 

Drawn  off  for  combustion  15.8  c.c. 

Volume  after  adding  oxygen  48.0  c.c. 
*'  **         '*         air  79.7  c.c. 

"  **     explosion  53.5  c.c. 

"  **     washing  with  KOH  46.0  c.c. 

The  volume  of  CH4  =  COo  formed  =  53-5  —  46  = 
7.5  c.c. 

As  the  volume  of  hydrogen  =  ^  of  contraction  — 
twice  the  CO2  formed:  we  have, 

Volume  of  hydrogen  =  ^  (26.2  —  15)  =  7.46  c.c. 

To  refer  these  quantities  to  the  original  100  c.c.  we 
have,  residual  volume  =  100  -  14.2  c.c.  =  85.8  c.c.  and 
for  H2,  15.8  :  85.8  :  :  7.46  :  :  x.     x  =  40.5 
for  CH4,  15.8  :  85.8  :  :     7-5  •:  Y-     Y  =  40.7 

So  our  results  become  :  carbonic  acid,  none :  illum- 
inants, 6.5%:  oxygen,  (7.4  —  6.5)0.9%:  carbonic  oxide, 
(14.2-7.4)  6.8%:  hydrogen,  40.5%:  methane  &c.; 
40.7%:  nitrogen  (by  difference),  4.6%. 


EXAMPLES. 


1.  Reduce  twelvelitersof  oxygen,  measured  at  47°C 
and  under  a  pressure  of  782  m.m.  of  mercury,  to  the 
standard  conditions.  Ans.      10.5338  liters. 

2.  What  will  be  the  volume  of  nine  liters  of  hydro- 
gen (at  the  standard  conditions)  when  heated  to  i2o°C 
and  under  a  pressure  of  423  m.m.  ? 

Ans.     23.278  liters. 

3.  Given  27  c.c.  of  nitrogen  at  i4°C  and  752.1  m.m. 
What  will  be  its  volume  at  20°C  and  767.8  m.m.  ? 

Ans,     27  c.c. 

4.  What  is  the  weight  of  77.2  c.c.  of  air  measured 
over  water  at  2  7°C  and  758  m.m.  pressure?  See 
tables.  •  Ans.     0.08742  gram. 

5.  What  is  the  weight  of  17  liters  of  hydrogen, 
measured  at  I2°C  and  3  atmospheres  pressure?  Take 
I  atmosphere  =  760  m.m  of  mercury. 

Ans.     4-38354  grams. 

6.  What  is  the  weight  of  85  c.c.  of  NH3  gas  meas- 
ured over  mercury  at  i40°C  and  under  a  pressure  of  770 
m.m.  ?  Given  the  tension  of  mercury  vapor  at  i40°C  = 
1.76  m.m.  Ans.     0.04327  gram. 

7.  Find  the  density  referred  to  air  of  chloroform 
vapor.      Given  the  following  data  for  Dumas'  method. 

Weight  of  bulb  and  air  at  20.5°C  and  763.9  m.m. 
34.8451  grams. 


122  EXAMPLES. 

Weight  of  bulb  and  vapor  at  ii6.5°C  and  763.9 
m.m.  35.8430  grams. 

Volume  of  air  in  bulb  at  2o.5°C  394.93  c.c. 

Ans.     4.10 

8.  What  is  the  density  referred  to  hydrogen  of  a 
substance  which  gave  the  following  results  by  Victor 
Meyer's  method  ? 

Weight  of  substance  taken  0.1307  gram. 

Volume  of  air  expelled  (measured  over  water)  25.8  c.c. 

Temperature  of  air  expelled  20°  C. 

Height  of  barometer  763.9  m.m.         Ans.     61.67. 

9.  What  is  the  molecular  weight  of  a  substance 
which  gave  the  following  results  by  Hofmann's  method. 

Weight  of  substance  used  0.1002  gram. 
Volume  of  vapor  62.35  c.c. 
Height  of  barometer  at  2i.5°C  771.5  m.m. 
Height  of  column  of  mercury — (inside  heating  jacket) 
at  ioo°C  314  m.m. 

Height  of  column  of  mercury — (outside  heating  jacket) 
at  26.5°C  135  m.m. 

Tension  of  mercury  vapor  at  ioo°C  0.27  m.m. 

Ans.     114.27. 

10.  What  is  the  density  referred  to  air  of  a  gas, 
which  escapes  in  14.5  minutes  through  an  opening, 
through  which,  at  the  same  temperature  and  pressure, 
the  same  volume  of  air  requires  19.5  minutes  ? 

Ans.     0.55. 

11.  What  is  the  density  referred  to  hydrogen  of  a 
gas,  which  escapes  in  7  minutes  30  seconds  from  an 
opening,  while  the  same  volume  of  air  under  the  same 
conditions  requires  5  minutes  10  seconds? 

Ans.     30.38. 


EXAMPLES.  123 

1  2.  An  object  whose  specific  gravity  is  3.38  (HoO  = 
I.),  weighed  39.7250  grams  in  air,  when  weighed  with 
brass  weights.     What  is  its  true  weight? 

Ans.     39-7335  grams. 

13.  An  object  whose  specific  gravity  is  0.95  weighed 
i7.8540grams  in  air,  when  weighed  with  brass  weights. 
What  is  its  true  weight?  Ans,     17.8740  grams. 

14.  A  bar  of  gold  Sp.  Gr.  19.3  weighed  10.73685 
kilos,  when  weighed  in  air  with  brass  weights.  What 
is  the  true  weight  ?  Ans,     10.73600  Kilos. 

15.  A  piece  of  pyrrhotite,  Sp.  Gr.  4.5,  weighed  in  air 
47.3854  grams,  when  weighed  with  platinum  weights. 
What  is  its  true  weight  ?  Ans.     47.3954  grams. 

16.  A  piece  of  antimony,  Sp.  Gr.  6.8,  weighed  75 
grams  in  air;  the  weights  used  were  50  grams  brass,  20 
and  5  grams  platinum.     What  is  the  true  weight  ? 

Ans.     75.0048. 

17.  If  50  c.c.  of  CH4,  50  c.c.  of  Ho  and  50C.C.  of 
C2H4  were  mixed  with  450  c.c.  of  oxygen  and  exploded. 
What  would  be  the  resulting  volume  at  the  .same  tem- 
perature and  pressure  ?  (Assume  that  the  temperature 
is  below  ioo°C.).  Ans.     325  c.c. 

18.  Calculate  the  percentage  by  volume  of  a  blast 
furnace  gas,  giving  the  following  results: 

Taken  for  analysis  100  c.c. 
Volume,  after  washing  with  KOH  (COo)  99.4  c.c. 

<'  "  <*         "     Br  (illuminants)  99.4C.C. 

<'  "  *'         **  Alkaline  **pyro"  (Oo)99.4c.c. 

-  CuoCL,HCl  (CO)  65.1  c.c. 

Decrease  in  volume  after  explosion 

(the  whole  used)  2.1  c.c. 

Removed  by  washing  with  KOH  none. 
Ans.     CO,  0.6%.     0034.3%-     H,  1.4%.     N^63.7%. 


124  EXAMPLES. 

19.  Calculate  the  percentage  by  volume  of  a  pro- 
ducer gas  giving  the  following  results  : 

Volume  taken  for  analysis  100  c.c. 

after  washing  with  KOH  (CO2)  98.5  c.c. 

''    Br(illuminants)  98.5  c.c. 
''    Alk.  *'pyro."  (Oo)  98.5  c.c. 
''   CugClo,  HCl  (CO)   75.0  c.c. 
Decrease  in  volume  after  explosion,  the  whole  used, 
15  c.c.     Removed  by  washing  with  KOH  3  c.c. 

Ans.  CO,,  1.5%.  CO,  23.5%.  CH„  3%.  H^,  6%. 
Ng  by  difference,  66%. 

20.  Calculate  the  composition  of  a  coal  gas  given  the 
following  data : 

Volume  taken  for  analysis  100  c.c. 

"       after  washing  with  KOH  (COo)  99.5  c.c. 

"  **         ''     Br(illuminants)  95.5  c.c. 

"     Alk.  ''pyro."(Oo)  95.0  c.c. 
-      CugClo,  HCl(cb)  89.OC.C. 
Decrease  after  explosion,  one  quarter  used,  36.9  c.c. 
Removed  by  washing  with  KOH  10.  i  c.c. 
Ans,     CO2,  0.5%.   Illuminants  (C0H4  &c.),  4%.   O2, 
0.5%.    CO,  6%.    CH4&C.,  40.4%.    H2,44.5%.    N2,by 
difference,  4.1%. 

2 1 .  Calculate  the  percentage  composition  of  an  illumi 
nating  gas  from  the  following  data.     Volume  taken  for 
analysis  too  c.c. 

Volume  absorbed  by  KOH      .         .  3.2  c.c. 

'*  ''  ''    Bromine  .  10.7  c.c. 

"  "  '*   alkaline  pyrogallol     2.7  c.c. 

"  "  ''    cuprous  chloride      27.4  c.c. 

To  16  c.c.  of  the  residual  gas  64  c.c.  of  air  were  added 
and  the  mixture  exploded.  The  contraction  was  21.4 
c.c.     Washing  with  KOH  solution  removed  4.9  c.c. 


EXAMPLES.  125 

Ans.  CO2  3.2%.  Illuminants  10.7%.  Og  2.7%. 
CO  27.4%.     CH,   17.15%.     H^  27.05%.     N,  11.8%. 

22.  Calculate  the  composition  of  an  illuminating  gas 
given  the  following  data.  Volume  taken  for  analysis  100 
c.c.  Volume  after  absorbing  with  absolute  alcohol  (hydro- 
carbons) 95.8  c.c.  Volume  after  absorbing  with  KOH 
(CO 2)  93.2  c.c.  Volume  after  absorbing  with  fuming 
sulphuric  acid,  and  then  with  KOH  (illuminants)  85.8  c.c. 
Volume  after  absorbing  with  alkahne  pyrogallol  (O2) 
82  c.c.  Volume  after  absorbing  with  cuprous  chloride 
(CO)  61.6  c.c. 

16  c.c.  of  the  residual  gas  was  diluted  to  80  c.c.  with 
air  and  exploded. 

Contraction  after  explosion  18.4  c.c.  Further  con- 
traction on  washing  with  KOH  3.5  c.c. 

Ans.  Hydrocarbons  4.2%.  CO 2  2.6%.  Illumi- 
nants 7.4%.  O2  3-8%.  CO  20.4%.  CH4  13.5%. 
H2  29.3%.     N  2  18.8%. 

23.  Find  the  molecular  weight  of  bromine.  Given 
the  following  data  for  Dumas'  Method.  Weight  of  bulb 
and  air  at  760.7  m.m.  and  24.5°  C.  51.8384  grams. 

Weight  of  bulb  and  bromine  vapor  at  760.7  m.m.  and 
102°  C.  53.1338  grams. 

Weight  of  bulb  and  water  at  24.5"^  C.  371.3  grams. 

To  calculate  the  volume  of  the  bulb  y. 

Let  X  =  weight  of  bulb,  then,  51.8384  —  (y  X  wt.  i 
c.c.  air)  =  x. 

Also  371.3  —  (yXwt.  I  c.c.  water)  =  x. 

So  51.8384  -  (yX. 0012)  =  371. 3 -(yXo.9972).  y 
=  320.7  c.c. 

Taking  the  gram  molecular  volume  as  22.4  liters  the 
molecular  weight  of  bromine  is  Ans.     160.67. 


126  EXAMPLES. 

24.  Given  the  weight  of  a  bulb  filled  with  air  at  24° 
C.  and  761  m.m.  pressure  50.178  grams,  the  weight  of 
the  same  bulb  filled  with  water  at  24°  C.  355.6  grams. 
What  is  the  weight  and  the  volume  of  the  bulb  ? 

Ans.     49.8131  grams.     306.6  c.c. 

25.  What  is  the  volume  of  the  products  of  the  com- 
plete combustion  of  5  grams  of  benzene  C^Hg,  when 
measured  at  70°  C.  and  760  m.m.  pressure?  Given  the 
weight  of  a  liter  of  hydrogen  0.08973  gram. 

Ans.     10.85  liters. 

26.  What  is  the  volume  occupied  by  the  products  of 
the  complete  combustion  of  10  grams  of  95%  CH3OH 
(the  remaining  5%  being  water)  at  4°  C.  and  768  m.m, 
pressure?     One  liter  of  hydrogen  weighs  0.08973  gram. 

Ans,     6.70  liters. 

27.  What  is  the  volume  occupied  by  the  products  of 
the  complete  combustion  of  1 2  grams  of  absolute  alcohol 
at  1 20°  C.  and  742  m.m.  pressure  ?  Given  the  weight 
of  one  liter  of  hydrogen  0.08973  gram. 

Ans.     43.165  liters. 

28.  What  is  the  volume  of  the  water  produced  by  the 
complete  combustion  of  13  liters  of  acetylene  measured 
under  the  standard  conditions  ?  Given  the  weight  of  one 
liter  of  hydrogen  0.08973  gram.  Ans.     10.4  c.c. 

29.  Calculate  the  volume  of  the  products  of  the  com- 
plete combustion  of  2  grams  of  glucose  C6Hi206ati7°C. 
and  758  m.m.  pressure.  Given  only  the  weight  of  a  liter 
of  hydrogen  0.08973  gram.  Ans.     1.59  liters. 

30.  Calculate  the  volume  occupied  at  105°  C.  and 
760  m.m.  by  the  products  of  the  complete  combustion  of 
100  c.c.  of  a  gas,  measured  at  0°  C,  containing  50% 
hydrogen,  40%  CO,  5%  CO 2  and  5%  N2  (by  volume). 

Ans.     138.4  c.c. 


CHAPTER  IX. 

CALCULATIONS   OF   CALORIFIC   POWER. 

Before  taking  up  the  calculations  of  the  calorific 
power  of  the  usual  fuels,  a  few  words  on  thermo-chemis- 
try  are  necessary,  to  serve  as  an  introduction  to  this  and 
the  following  chapter.  We  have  already  discussed  the 
qualitative  and  quantitative  aspects  of  a  chemical  re- 
action ;  it  remains  to  show  here  the  thermo-chemical 
relations  expressed. 

All  chemical  reactions  are  accompanied  by  the  liber- 
ation or  absorption  of  a  definite  amount  of  energy,  which 
may  manifest  itself  in  various  ways,  such  as  heat  or  elec- 
tricity.    Let  us  consider  the  equation  : 

Zn  +  CI2  =  ZnCl2,  as  written  this  signifies  that  65.4 
parts  by  weight  of  zinc  combine  with  70.9  parts  of 
chlorine  to  give  136.3  parts  of  zinc  chloride  ;  but  besides 
this,  heat  is  liberated  equivalent  to  97,200  calories,  when 
the  quantities  used  are  the  atomic  or  molecular  weights 
in  grams,  so 

Zn  -j-  CI2  =  ZnClg  +  97,200  calories. 
In  other  words  65.4  grams  of  zinc  and  70.9  grams  of  chlo- 
rine contain  when  uncombined  potential  energy  equiva- 
lent to  97,200  calories  more  than  that  of  136.3  grams  of 
zinc  chloride  ;  consequently  to  decompose  zinc  chloride, 
into  zinc  and  chlorine,  energy  must  be  expended,  equiva- 
lent to  that  given  up  when  combination  took  place  or 
97,200  calories  per  gram-molecule.     This  energy  can  be 

127 


128  CALCULATIONS   OF   CALORIFIC    POWER. 

conveniently  supplied  in  the  form  of  an  electric  current ; 
and  the  calculation  of  the  voltage  necessary  to  decom- 
pose certain  molecules  will  be  discussed  in  the  next 
chapter. 

The  heats  of  formation  of  many  molecules  have  been 
determined  and  are  usually  expressed  in  terms  of  heat 
developed  by  the  formation  of  the  molecular  weight  in 
grams  of  the  substance  ;  so  that  the  numbers  are  directly 
comparable  with  the  molecules  taking  part  in  reactions. 
This  is  not  always  the  case  however  and  care  m.ust  be 
taken  to  ascertain  whether  the  table  is  based  on  gram- 
molecules,  gram-equivalents  or  grams  of  the  compound 
formed. 

The  unit  of  heat  is  the  calorie,  the  amount  of  heat 
necessary  to  raise  one  gram  of  water  from  4°  to  5°  C.  : 
as  the  specific  heat  of  water  does  not  vary  greatly  with 
the  temperature,  a  calorie  is  taken  practically  as  the 
amount  of  heat  necessary  to  raise  one  gram  of  water  one 
degree  centigrade.  Other  units  in  use  are :  the  greater 
calorie,  the  heat  necessary  to  raise  one  kilogram  of  water 
1°  C,  equal  to  1000  calories:  and  a  value  denoted  by 
K,  introduced  by  Ostwald,  the  amount  of  heat  necessary 
to  raise  one  gram  of  water  from  0°  to  100°  C,  this  is 
practically  equal  to  100  calories.* 

The  following  are  important  thermo-chemical  re- 
actions : 

H2  +  O  =  H.O  +  68400  calories  (684  K) 
C  +  O  =  CO  +  29000  ''  (290  K) 
CO  +  O  =  COo  +  68000  ''  (680  K) 
C  +  Oo  =  COo  +  97000  ^'  (970  K) 
S      +02=  SOo  +  71000      -         (710  K) 

*  A  very  full  table  of  heats  of  formation  expressed  in  terms  of  K  will  be  found 
in  Ostwald's  Outlines  of  General  Chemistry. 


CALCULATIONS   OF   CALORIFIC    POWER.  129 

We  see  from  these  the  amounts  of  heat  given  up  by  the 
formation  of  i8  grams  of  water,  28  grams  of  carbonic 
oxide  etc.  For  the  purpose  of  calculating  the  calorific 
power  of  fuel,  it  will  be  more  convenient  to  have  the 
amount  of  heat  formed  by  the  combustion  of  one  gram 
of  hydrogen,  carbon  and  sulphur.  If  2  +  grams  of  hydro- 
gen and  16  grams  of  oxygen  give  18  +  grams  of  water 
and  68,400  calories :    one   gram  of  hydrogen  will  give 

— — —  or  34,200  calories  :  similarly  one  gram  of  carbon 

21 or  8083  calories  and  one  e^ram  of  sulphur  7^^^^^ 

12  .  ^  32 

=  2219  calories. 

These  heats  of  formation,  so  essential  to  the  calcula- 
tion of  calorific  power,  are  given  various  figures  by  dif- 
ferent observers   and  it  is  a  difficult  matter  to  decide 
which  to  use.     We  shall  adopt  the  following  as  having 
the  greatest  probability  of  accuracy  and  at  the  same  time 
being  in  most  general  practical  use. 
One  gram  carbon  to  carbonic  acid  8080  calories 
One  gram  hydrogen  to  water  (condensed)  34500  calories 
One  gram  sulphur  to  sulphur  dioxide  2220  calories 
These  numbers  are  the  same  in  greater  calories  when 
one  kilogram  of  the  element  is  burned. 

It  will  be  seen  from  the  preceding  variations  between 
the  values  of  the  fundamental  data,  that  considerable  un- 
certainty exists  concerning  thermo-chemical  figures. 
The  relative  accuracy  is  much  greater  than  the  absolute; 
so  the  data  employed  should  always  be  taken  from  the 
same  source  and  if  possible  obtained  by  the  same 
method. 


130  CALORIFIC    POWER   OF   FUEL. 

With  this  introduction  we  proceed  to  the  consider- 
ation of  the  calculations  of  the  calorific  power  of  fuel 
from  a  practical  standpoint. 

CALORIFIC    POWER    OF    FUEL. 

The  calorific  power  of  any  solid  fuel  such  as  coal  can 
be  determined  by  three  methods. 

I  St.   By  combustion  in  a  calorimeter. 

2nd.  By  calculation,  from  the  results  of  an  ultimate 
analysis. 

3rd.  By  calculation,  from  the  weight  of  lead  reduced 
from  litharge  by  fusion  with  a  weighed  quantity  of  the 
coal. 

Only  the  first  and  second  methods  are  applicable  to 
liquids  or  gases. 

1st.  Of  these  three  methods  the  first  is  the  most  ac- 
curate, as  it  affords  a  direct  measure  of  the  heating 
power  of  the  fuel  when  burned  under  the  most  favorable 
conditions.  The  calculations  involved  do  not  specially 
belong  to  analytical  chemistry,  so  the  reader  is  referred 
to  Poole  on  the  Calorific  Power  of  Fuels,  for  both  a  de- 
scription of  the  apparatus  and  direciions  for  the  calcula- 
tions. 

2nd.  The  second  method  is  largely  used  and  is  im- 
portant for  analysts,  as  they  usually  have  the  apparatus 
required  to  get  the  analytical  results,  while  they  are  not 
often  supplied  with  an  expensive  calorimeter. 

For  commercial  purposes,  calorific  power  is  expressed 
in  four  ways  : 

1st.  In  calories  per  kilogram  (greater  calories),  for 
example  7000  calories  means  that  one  kilogram  of  the 
fuel  will  raise  the  temperature  of  7000  kilograms  of 
water  one  degree  centigrade. 


CALORIFIC    POWER   OF   FUEL.  131 

2nd.  In  British  thermal  units  (B.T.U.)  or  the  amount 
of  heat  necessary  to  raise  one  pound  of  water  one  degree 
Fahrenheit. 

3rd.  In  kilograms  of  water  evaporated  per  kilogram 
of  coal. 

4th.  In  pounds  of  water  evaporated  per  pound  of  coal 

The  relations  between  these  various  ways  of  express- 
ing calorific  power  are  not  complicated.  To  convert 
calories  per  kilogram  into  British  thermal  units  it  is  only 
necessary  to  multiply  by  1.8  for  if  a  kilogram  of  coal 
heats  7000  kilograms  of  water  one  degree  centigrade, 
one  pound  of  coal  will  heat  7000  pounds  of  water  one  de- 
gree centigrade,  and  it  only  remains  to  multiply  by  the 
factor  for  converting  centigrade  to  Fahrenheit  degrees. 
Conversely  to  change  B.T.U.  to  calories  divide  by  1.8 
or  multiply  by  %. 

To  find  the  number  of  kilograms  of  water  evaporated 
by  a  kilogram  of  fuel,  divide  the  calorific  power  (in 
calories  per  kilogram)  by  536,  the  number  of  calories  re- 
quired to  convert  one  kilogram  of  water  at  100°  C  into 
steam  at  100°  C.  If  it  should  be  necessary  to  raise  the 
water  from  some  temperature,  for  example  15°  C  to« 
100°  C  and  then  evaporate  it :  divide  by  536  +  (100  -  15) 
or  621.     As  the  specific  heat  of  water  is  taken  as  unity.. 

To  find  the  number  of  pounds  of  water  evaporated  by 
a  pound  of  fuel,  divide  the  calorific  power  expressed  in 
B.T.U.  by  965  (536  X  1.8).  For  example,  how  many 
pounds  of  water  will  be  heated  from  59°  F.  to  212°  F. 
and  evaporated  by  one  pound  of  a  coal  whose  calorific 
power  is  7000  (calories  per  kilogram)  ?  7000  X  1.8  = 
12,600  B.T.U.  It  requires  212-59  =  153  B.T.U. 
to  heat  a  pound  of  water  from  59°  F.  to  212°  F.  and 
965  B.T.U.  to  convert  it  into  steam  at  212°  F:  So, 
12600  ^  (153  +  965)  =  11.27  pounds. 


132  CALORIFIC    POWER    OF    SOLIDS. 

CALORIFIC    POWER    OF    SOLIDS. 

Given  the  composition  of  perfectly  dry  oak  wood : 
carbon,  50.16%:  hydrogen,  6.02%:  oxygen,  43.36%: 
nitrogen,  0.09%:  ash,  0.37%:  to  calculate  its  calorific 
power. 

One  kilogram  will  contain : 
Carbon       501.6  grams  Nitrogen  0.9  grams 

Hydrogen    60.2       ''  Ash  T,.y      *' 

Oxygen      433.6       '' 

As  part  of  the  hydrogen  is  already  in  combination 
with  oxygen,  only  the  excess  over  the  amount  necessary 
to  form  water  with  all  the  oxygen  is  available  as  fuel : 
as  16  parts  of  oxygen  combine  with  2  of  hydrogen,  the 
hydrogen  already  in  combination  with  oxygen  will  be 
433.6  -i-  8  =  54.2  grams:  So  we  have  available  as  fuel 
6  grams  of  hydrogen  and  501.6  grams  of  carbon. 

0.006  kilogram  hydrogen    X  345oo  =     207  calories 
0.5016        *'         carbon        X    8080  =  4053        " 

Theoretical  heating  power  =  4260  calories. 
If  the  products  of  combustion  escape  at  a  temperature 
of  or  above  ioo°C.,  the  heat  given  up  by  the  condensa- 
tion of  the  steam,  which  is  lost,  must  be  subtracted. 
This  is  found  as  follows  :  60.2  grams  of  hydrogen  give 
9  X  60.2  =  541.8  grams  of  water;  and  as  536  calories 
are  required  to  convert  one  kilogram  of  water  at  100°  to 
steam  at  100°,  the  heat  lost  in  this  case  will  be  equal  to 
that  required  to  change  541.8  grams  of  water  into  steam 
at  ioo°C  or  536  X  0.5418  =  290  calories  :  so  the  avail- 
able heat  becomes  3970  calories.  When  the  fuel  has 
also  hygroscopic  moisture  this  must  also  be  evaporated 
and  if  the  products  are  not  condensed,  the  heat  neces- 
sary to  do  this  must  be  deducted. 


CALORIFIC   POWER   OF   SOLIDS.  133 

To  calculate  the  calorific  power  of  a  coal  whose  com- 
position is:  carbon,  80.55%;  hydrogen,  4.50%;  sul- 
phur, 0.54%;  oxygen,  (by  difference)  2.70%;  nitrogen, 
1.08%;  moisture,  2.92%;  ash,  7.71%. 

The  nitrogen  and  ash  are  inert  and  simply  diminish 
the  available  calorific  power  by  the  heat  expended  in 
bringing  them  to  the  temperature  of  the  products  of  com- 
bustion, this  amount  is  relatively  very  small  and  is  neg- 
lected. 

The  moisture  is  objectionable  as  it  must  be  raised  to 
ioo°C  and  then  evaporated,  thus  rendering  latent  a  por- 
tion of  the  heat  evolved  which  can  however  be  recovered 
if  the  products  of  combustion  are  cooled  below  ioo°C. 

The  available  hydrogen  is  calculated  as  before, 

4-5o--g^   =  4.16%. 

One  kilogram  of  the  coal  contains  available  as  fuel : 
41.6  grams  of  hydrogen  giving  1435.2  calories 
805.5        ''       ''  carbon  ''       6508.4  *' 

5.4        ''       "  sulphur         "  12.0         *^ 

Theoretical  calorific  power  =  7943.6         " 

This  on  the  basis  of  the  condensation  of  the  aqueous 
vapor  produced  which  will  then  give  up  heat  equal  to 
that  required  to  vaporize  it,  so  no  deduction  is  made.  If 
the  products  are  allowed  to  escape  without  condensation 
of  the  aqueous  vapor,  as  is  always  the  case  in  practice ; 
the  number  of  calories  necessary  to  convert  into  vapor 
all  the  water  present  must  be  subtracted. 

This  will  be,  9  X  4-5  (the  percentage  of  hydrogen)  = 
405  grams  per  kilogram  plus  moisture  29.2  grams  = 
434.2  grams  or  0.4342  kilogram;  and  0.4342  X  53^ 
(the  number  of  calories  necessary  to  convert  one  kilo- 


134  CALORIFIC   POWER   OF   SOLIDS. 

gram  of  water  at  ioo°C  to  steam  at  ioo°C)  gives  232.7 
calories.  So  under  these  conditions  the  available  calori- 
fic power  of  the  coal  is  7710.9  calories.  This  can  be 
corrected  further  for  the  heat  necessary  to  raise  the  fuel 
to  ioo°C  and  for  the  heat  lost  by  the  escape  of  the  pro- 
ducts of  combustion  at  temperatures  above  ioo°C  &c. 

Taking  771 1  as  the  calorific  power  of  the  coal,  let 
us  see  the  quantities  of  water  it  will  evaporate. 

Kilograms  of  water  at  ioo°C  per  kilogram  of  coal 
will  be  771 1    ^  14.38. 

Kilograms  of  water  at  I5°C   per  kilogram  of  coal 

will  be  ^A r-  =  12.41. 

536  +  (100-15) 

Pounds  of  water  at  2i2°F.  per  pound  of  coal 
will  be  77iL^i_L8  ^,   TZii  ^         g. 
965  536 

Pounds  of  water  at  4o°F.  per  pound  of  coal 

will  be  — ^-^ — X  =  12.21. 

965  +  (212-40; 

This  discussion  can  be  summed  up  by  the  following 

formulae.      Let  x  be  the  calorific  power  of  the  fuel,  and 

let  C,H,0,S  and  W  be  the  weights  of  carbon,  hydrogen, 

oxygen,    sulphur   and  hygroscopic  water,  expressed  in 

kilograms:   then  we  have, 

X  =  8080C  +  34500  (H  -  9)  +  2220  S. 

o 

An  expression  for  the  theoretical  calorific  power.  And, 
x  =  8o8o  C  + 34500  (H-§)+222oS-536(W  +  9H) 

o 

For  the  available  calorific  power  when  the  products 
of  combustion  are  not  condensed.      The  heat  generated 


CALORIFIC   POWER   OF   LIQUIDS.  135 

by  the  sulphur  can  usually  be  omitted  without  materi- 
ally affecting  the  results. 

Numerous  formulae  are  given  by  different  authori- 
ties, which  differ  somewhat  from  the  preceding  and 
attempt  to  bring  the  results  by  calculation  nearer  those 
obtained  by  the  calorimeter,  or  the  conditions  of  actual 
practice.  Those  given  agree  very  closely  with  that  of 
Dulong  as  recommended  by  the  committee  of  the  Amer- 
ican Chemical  Society.* 

This  method  of  calculation  is  based  on  Welter's  rule, 
where  the  assumption  is  made  that  the  oxygen  present  is 
all  combined  with  hydrogen  or  that  we  shall  offset  the 
heat  of  formation  of  the  molecules  present  by  deducting 
the  quantity  of  heat  which  would  be  produced  if  the 
oxygen  present  combined  with  hydrogen.  This  is  a  very 
weak  point  in  the  method  of  obtaining  the  calorific  power 
from  the  percentage  composition. 

In  the  case  of  anthracite  the  error  will  be  small 
because  the  combustible  material  is  nearly  all  carbon. 
With  a  bituminous  coal  the  error  is  liable  to  be  much 
larger,  while  with  wood  the  attempt  to  offset  the  unknown 
heat  of  combination  of  cellulose  and  the  other  complex 
organic  compounds  present  by  this  deduction  is  very 
unsatisfactory. 

Ostwald  has  recommended  subtracting  the  oxygen 
present  from  the  carbon  instead  of  from  the  hydrogen  as 
giving  results  closer  to  those  obtained  by  actual  com- 
bustion, f 

CALORIFIC  POWER  OF  LIQUIDS. 

The  calculation  can  be  made  as  already  given  but  the 
results  are  necessarily  unsatisfactory  unless  the  heat  of 

*J.  Am.  Chem.  Soc.,  Dec,  1899. 

t  For  this  method  of  calculation,  see  Walker,  Introduction  to  Physical  Chemistry, 
p.  129. 


136  CALORIFIC    POWER   OF  GASES. 

combination  is  considered,  which  is  not  readily  obtained 
for  mixtures  like  crude  petroleum,  etc.  For  alcohol  the 
calculation  could  be  made,  but  for  all  commercial  liquid 
fuels  the  determination  of  calorific  power  should  be  made 
by  the  calorimeter. 

CALORIFIC  POWER  OF  GASES. 

In  connection  with  the  calorific  power  of  gases  the 

discrepancy  in  results  between  the  calculated  and  actual 

calorific  power  of  solids  and  liquids  is  best  explained. 

Let  us  consider  the  thermo-chemical  reactions  : 

C  +   O2  =  CO2     +  970  K 

2H2+    Oo  =  2H2O  +  2(684)K 

C  +  2H2  =  CH4     +  218  K  hence 
CH4+2O,  =  COo     +2H0O+2120K 
2 1 20  =  970  +  2(684)  -  2 1 8, 
or  the  heat  produced  by  the  combustion  of  16  grams  of 
CH4is  2120  K;   less  by  218  K  than  the  heat  produced 
by  the  burning  of  1 2  grams  of  carbon  and  four  grams  of 
hydrogen. 

In  solids  and  liquids  when  we  do  not  know  the  chemi- 
cal compounds  and  only  the  ultimate  composition  we  may 
be  led  into  similar  errors. 

Sixteen  grams  of  methane  (CH4)  gave  2120  K  hence 
one  gram  gives  132.5  K  or  13250  calories  and  one  kilo- 
gram the  same  number  of  greater  calories.  In  the  table 
at  the  back  of  this  book,  we  find  the  weight  of  a  liter  of 
methane  to  be  0.71502  gram.  The  weight  of  a  cubic 
meter  is  1000  times  this  or  an  equal  number  of  kilograms. 
The  calorific  power  of  a  cubic  meter  will  therefore  be 
13250  X  0.715  =  9474  calories  (greater),  and  as  a  cubic 
meter  =  35.3171  cubic  feet,  the  calorific  power  per  cubic 
foot  is   268.2  calories.     To  convert  this  to  B.T.U.  per 


CALORIFIC    POWER    OF   GASES.  137 

cubic  foot  it  must  be  multipled  by  3.97  giving    1065  as 
the  calorific  power  of  methane  per  cubic  foot. 

Now  that  we  see  the  relation  of  these  values  we  can 
conveniently  use  the  following  figures  to  calculate  the 
theoretical  calorific  power  of  a  gas  per  cubic  foot  meas- 
ured at  0°  C  and  760  m.m.  pressure 

Hydrogen,  H2        347.2    B.T.U  per  cubic  foot. 

Carbonic  oxide  CO.      341.4        "         "       "        '* 

Methane  CH4     1065         ''         *^ 

llluminants  C0H4  +  C.^Hg  2000  B.T.U.  per  cubic  foot. 
If  we  assume  the  initial  temperature  of  the  gas  to  be 
60°  F.  and  the  products  to  escape  at  328°  F.,  without 
condensation,  these  figures  become  :  * 

Hydrogen  Ho        264.      B.T.U.  per  cubic  foot 

Carbonic  oxide  CO       307.  *'         "       **         " 

Methane  CH4     853. 

llluminants  C0H4+C3H6  1700  " 
Given  a  gas  of  the  following  composition,  to  calculate 
its  theoretical  calorific  power  in  B.T.U.  per  cubic  foot: 
hydrogen,  50.4%:  methane,  33-7%'  illuminants, 
5.2%:  carbonic  acid,  2.6%:  carbonic  oxide,  5.8%: 
oxygen,  0.8%:   nitrogen,  1.5%. 

As  all  gases  expand  equally  the  percentage  compo- 
sition by  volume  is  not  affected  by  the  temperature  at 
which  the  analysis  was  made.  In  this  case  the  oxygen 
cannot  be  in  combination  with  the  hydrogen,  so  we  have 
available  as  fuel : 

Hydrogen    0.504  Cu.  ft.  X  347-2  =  174-98  B.T.U. 

Methane      0.337         ''     X  1065.  =  358.90 

Illuminants  0.052         **     X  2000    =104.00       *' 


*  For  details  of  these  calculations  see   Stillman's   Engineering  Chemistry, 
p.  263. 


138      CALORIFIC  POWER  FROM  WEIGHT  OF  LEAD  REDUCED. 

Carbonic  oxide  0.058  Cu.  ft.  X  341.4  =  19.80  B.T.U. 

Theoretical  calorific  power  per  Cu.  ft.  657.68  B.T.U. 
To  convert  this  to  calories  per  cubic  meter  multiply  by 
8.896:  which  gives  5851. 

The  available  calorific  power,  under  the  conditions 
stated,  is  found  by  using  the  values  in  the  second  table. 

Many  corrections  might  be  added  and  other  calcu- 
lations, such  as  the  volume  of  air  required,  the  volume  of 
the  products  of  combustion,  calorific  intensity  &c.  Some 
of  these  can  be  done  by  the  methods  given  under  the 
calculations  of  gases,  the  others  seem  out  of  the  sphere 
of  analytical  chemistry. 

CALORIFIC    POWER    FROM    WEIGHT    OF    LEAD    REDUCED. 

The  method  of  calculating  calorific  power  from  the 
weight  of  lead  reduced  from  litharge  by  a  gram  of  the 
fuel  is  only  suitable  for  coal  or  coke  and  is  only  valuable 
when  the  average  weight  of  three  lead  buttons  is  multi- 
plied by  a  factor  proper  for  the  fuel. 

This  factor  should  be  obtained  by  comparison  with 
the  calorimeter  and  will  be  the  calories  developed  by 
one  gram  divided  by  the  lead  reduced  from  litharge  by 
one  gram.  When  this  is  once  obtained  it  will  give  sat- 
isfactory results  with  similar  coals. 

Berthier's  idea  that  the  calorific  power  was  directly 
proportional  to  the  oxygen  consumed,  i,  e.  the  lead  re- 
duced, is  disproved  by  the  following  : 

C+  2  PbO  =  2  Pb+  CO2 

12  parts  of  carbon  reduce  414  parts  of  lead  hence, 
I  kilogram  of  carbon  reduces  34.5  kilograms  of  lead. 
I  kilogram  of  carbon  gives  8080  calories,  hence, 

the  lead  reduced  times— ?—^  or  234.2  gives  the  calorific 

34.5 


CALORIFIC  POWER  FROM  WEIGHT  OF  LEAD  REDUCED.      139 

power.     The  factor  for  hydrogen  is  obtained  similarly : 
Ho+  PbO  =  Pb  +  R,0. 
I  kilogram  of  hydrogen  would  reduce  103.5  kilograms 
of  lead,  and  as  one  kilogram  gives  34500  calories,  the 

factor  is   5z5 — or  ^^^.^^. 
103.5 

So  it  is  evident  that  the  more  volatile  matter,  hydro- 
carbons &c.,  the  coal  contains,  the  higher  must  be  the 
factor  used. 

When  no  better  means  of  determining  calorific  power 
is  available  the  weight  of  lead  reduced  by  one  gram  times 
240  for  an  anthracite  or  times  26813*  ^^ra  bituminous 
coal,  will  give  approximately  the  calorific  power. 

For  a  discussion  of  the  calculations  required  in  using 
the  bomb  calorimeter  the  student  is  referred  to  an  article 
by  Professor  Atwater  and  Dr.  Snell  on  a  "Description 
of  a  Bomb  Calorimeter  and  Method  of  its  Use"  pubHshed 
in  the  Journal  of  the  American  Chemical  Society,  July, 
1903,  Vol.  XXV,  p.  659-699. 

*Noyes,  J.  Am.  Chem.  Soc,  Vol.  17,  848. 


EXAMPLES. 


1.  Find  the  theoretical  calorific  power  of  wood,  whose 
composition  is:  carbon,  40%;  hydrogen,  4.8% ;  oxygen, 
33'S%;  nitrogen,  0.4% ;  ash,  1%;  hygroscopic  water, 
20%.  Assuming  the  wood  to  be  at  ioo°C  and  that 
the  aqueous  vapor  is  condensed. 

Ans.     3430.4  calories. 

2.  Find  the  calorific  power  of  peat,  whose  composi- 
tion is:  carbon,  39.3%;  hydrogen,  3.9%;  oxygen,  23.2 
%;  ash,  7.3%;  water,  26.3%.  Assuming  that  the  pro- 
ducts of  combustion  escape  at  ioo°C.  without  conden- 
sation. Ans.     3 19 1. 3  calories. 

3.  Find  the  theoretical  calorific  power  of  a  coal, 
whose  composition  is  :  carbon,  83.75%;  hydrogen,  4. 13 
%;  oxygen.  2.65%;  nitrogen,  0.85%;  sulphur,  0.57%; 
water,  0.80%;  ash,  7.25%.       Ans.     8090.6  calories. 

4.  In  the  preceding  example,  how  many  calories 
must  be  deducted  if  the  aqueous  vapor  is  not  condensed? 

Ans.     203.5. 

5.  Find  the  theoretical  calorific  power  in  B.T.U.  of 
a  coal  whose  composition  is :  carbon,  70.50%;  hydro- 
gen, 4.76%;  oxygen,  15.71%;  nitrogen.  1.36%;  sul- 
phur, 1.39%;  ash,  6.28%.  A7ts.      12048. 

6.  How  many  kilograms  of  water  at  20°C  will  be 
evaporated  by  one  kilogram  of  a  coal,  whose  calorific 
power  is  13535  B.T.U.  ?  Ans.      12.2 

(140) 


EXAMPLES,  141 

7.  How  many  pounds  of  water  at  25°C  will  be  eva- 
porated by  a  coal  whose  calorific  power  is  8000  ? 

Ans.      1 3. 1 

8.  How  many  kilograms  of  water  at  6o°C  will  be 
evaporated  by  a  coal  whose  calorific  power  is  11 270 
B.T.U?  Ans.   10.86 

9.  What  is  the  difference  between  the  calculated 
available  calorific  powers  of  ethyl  and  methyl  alcohol. 

Ans,      2000  calories  (in  round  numbers). 

10.  What  is  the  calorific  power  of  a  natural  gas, 
whose  composition  is :  methane,  96.50%;  illuminants, 
i%\  carbonic  oxide,  0.5%;  oxygen,  2% ;  in  B.T.U. 
per  cubic  foot?     Gas  and  products  at  o°C. 

Ans.     1049. 

11.  What  is-  the  calorific  power  of  a  coal  gas,  whose 
composition  is :  hydrogen,  50.4%;  methane,  2>?>-7%'y 
illuminants,  5.2%;  carbonic  acid,  2.6%;  carbonic  oxide, 
5.8%;  oxygen,  0.8%;  nitrogen,  1.5%;  in  B.T.U.  per 
cubic  foot  ?  Assume  that  the  gas  enters  at  6o°F  and  the 
products  of  combustion  escape  at328°F. 

Ans.     52  7. 

12.  What  is  the  calorific  power  of  the  following 
water  gas:  Hydrogen,  55%;  methane,  5.75%;  illumin- 
ants, 1.20%;  carbonic  acid,  4.05%;  carbonic  oxide, 
31.00%;  nitrogen,  3%;  in  calories  per  cubic  meter? 
Gas  and  products  at  o°C.  Ans     3398. 


CHAPTER  X. 

ELECTRIC     AND    ELECTROLYTIC    CALCULATIONS    FOR 
DIRECT    CURRENTS. 

The  calculations  described  In  this  chapter,  although 
not  essential  to  analytical  chemistry,  are  added  because 
electricity  is  so  largely  employed  for  both  laboratory 
and  industrial  operations,  and  it  Is  advisable  that  a 
chemist  be  able  to  make  the  ordinary  calculations  invol- 
ved In  its  use.  In  order  to  make  the  description  as  con- 
cise as  possible  the  subject  is  treated  differently  from  the 
preceding  chapters,  and  the  Important  laws  are  expres- 
sed by  formulae,  many  of  which  are  taken  from  an  article 
by  Prof  Crocker  on  the  Theory  of  Electro- metallurgy.* 

ELECTRIC    CALCULATIONS. 

The  ampere  Is  the  unit  of  current,  that  is  the  rate  of 
flow ;  it  is  the  strength  of  current  w^hlch  when  passed 
through  a  solution  of  silver  nitrate  in  water,  under  cer- 
tain standard  conditions,  will  deposit  silver  at  the  rate  of 
0.001118  gram  per  second.  Five  amperes  will  deposit 
five  times  this  amount  eve^y  second.  So  to  deposit 
1. 34 1 6  grams  of  silver  from  an  aqueous  solution  of  silver 
nitrate  in  ten  minutes  requires  two  amperes ;   for 

0.001118  X  600 
The  coulomb  or  ampere-second  is  the  unit  of  quan- 
tity.    It  is  the  quantity  of  electricity  which  flows  along  a 

*  School  of  Mines  Quarterly.     January,  1895. 

(142) 


ELECTRIC   CALCULATIONS.  143 

conductor  in  one  second  when  the  current  is  one  ampere. 
The  total  quantity  or  number  of  coulombs  flowing  in  a 
given  time  is  equal  to  the  strength  of  current  in  amperes 
multiplied  by  the  time  in  seconds  or  expressed  in 
symbols 

Q  =  ct  (i) 

when  Q  =  quantity  in  coulombs 

C   =  strength  of  current  in  amperes 
t  =  time  in  seconds 

In  dealing  with  large  quantities  of  electricity  the 
ampere-hour  is  used  as  a  unit,  an  ampere  flowing  for  an 
hour.     This  is  3600  coulombs. 

If,  for  example,  1.5  amperes  flow  for  45  minutes  the 
number  of  coulombs  that  have  passed  will  be 
^•5  X  45  X  60  =  4050  equal  to  1.125  ampere-hours. 

The  oh7n  is  the  unit  of  resistance.  It  is  the  resist- 
ance offered  to  an  unvarying  electric  current  by  a  col- 
umn of  mercury  at  o°C,  one  square  millimeter  in  cross 
section  and  106.3  centimeters  long. 

The  resistance  of  a  conductor,  whether  solid  or 
liquid,  to  the  passage  of  electricity  has  been  found  to  de- 
pend upon  its  dimensions  and  the  material  of  which  it 
consists.  For  a  given  material  it  varies  directly  as  the 
length  and  inversely  as  the  area  of  the  cross  sec- 
tion. 

The  specific  resistance  of  any  material  is  the  resist- 
ance measured  between  opposite  faces  of  a  cube  of  the 
material  one  square  centimeter  in  area. 

The  resistance  of  any  conductor  is  given  by  the 
formula, 

R  =  P  1  (2) 


144  ELECTRIC   CALCULATIONS. 

in  which 

R  is  the  resistance  of  the  conductor  in  ohms 

p  is  the  specific  resistance  of  the  material  in  ohms 

1    is  the  length  in  centimeters,  and 

a    is  the  area  of  the  cross-section  in  square  centi- 
meters. 

Therefore,  if  the  specific  resistance  of  a  conductor  is 
known  the  total  resistance  can  be  calculated  from  its  di- 
mensions. To  find  the  resistance  of  200  meters  of  Ger- 
man-silver wire  having  a  cross  section  of  10  square 
millimeters;  given  the  specific  resistance  of  German- 
silver  as  0.0000209.*  Reduce  the  area  to  square  centi- 
meters and  the  length  to  centimeters  and  the  resistance 

v/  20000  O  1 

is  0.0000209  X  =4.18  ohms. 

0.1 

Similarly,  if  a  glass  tube,  20  centimeters  long  with 

an  internal  cross  section  of  2  square  centimeters,  is  filled 

with  a  solution  of  zinc  sulphate  whose  specific  resistance 

is  30  ohms,  the  total  resistance  will  be 

30  X  —  =  300  ohms. 
2 

The  specific  resistance  is  not  absolutely  constant  but 
changes  with  the  temperature.  In  the  case  of  all  metals 
it  increases  with  the  temperature  while  for  carbon,  selen- 
ium, liquids  and  solutions  it  diminishes  as  the  tempera- 
ture rises.  For  aqueous  solutions  it  naturally  varies 
with  their  strength. 

Conductors  are  connected  in  series  when  the  current 
passes  through  one  after  the  other;  in  parallel,  when 
they  are  connected  side  by  side  so  as  to  furnish  more 
than  one  path  for  the  current.  The  total  resistance  of  a 
number  of  separate  resistances  connected  in   series  is 

*  See  Tables. 


ELECTRIC  CALCULATIONS.  145 

equal  to  their  sum.  If  two  paths  of  equal  resistance  arc 
offered,  the  total  resistance  is  only  one  half,  as  this  is 
equivalent  to  doubling  the  cross  section  of  the  conduc- 
tor, or  the  resistance  of  a  number  of  equal  resistances  in 
parallel  is  equal  to  the  resistance  of  one  divided  by  the 
number  so  connected. 

Conductance  is  the  reciprocal  of  resistance,  or  the 
ease  with  which  a  current  is  allowed  to  pass.  The  total 
conductance  of  a  number  of  separate  conductances  con- 
nected in  parallel  is  equal  to  their  sum.  Thus  five  con- 
ductors each  of  50  ohms  resistance,  connected  in  series 
give  a  resistance  equal  to  their  sum,  250  ohms;  while 
connected  in  parallel  they  give  the  resistance  of  one 
divided  by  the  number,  10  ohms.  If  three  resistances  of 
20,  50  and  80  ohms  respectively,  are  connected  in  series 
their  total  resistance  is  150  ohms;  while  if  connected  in 
parallel  theirresistance  is  1 2. 1 2  ohms;  for  their  respective 

conductances  will  be  — '  —  and  —  and  their  total  con- 

20   50  bo 

ductance  the  sum  of  the  three,  0.0825.     So  that  their 

total  resistance  is  the  reciprocal  of  0.0825  or  12.12  ohms. 

The  vol^  is  the  unit  of  electromotive  force  or  elec- 
trical pressure. 

The  relations  between  electromotive  force,  current, 
and  resistance  are  expressed  by  Ohms  Law ;  that  the 
electromotive  force  acting  between  the  extremities  of  any 
part  of  a  circuit  is  the  product  of  the  strength  of  the  cur- 
rent  and  the  resistance  of  that  part  of  the  circuit,  or 
E  =  C  R 

or  transposed  C  =  —  w/ 

To  comply  with  Ohms  Law  and  the  definitions  of  the 
ampere  and  ohm  the  unit  of  electromotive  force,  the  volt 


146  ELECTRIC   CALCULATIONS. 

must  be  that  electromotive  force  which  steadily  applied 
to  a  conductor  whose  resistance  is  one  ohm  will  produce 
a  current  of  one  ampere. 

The  Clark  cell  at  15°  C  is  the  most  constant  source 
of  electromotive  force  and  is  used  as  a  standard.  It 
gives  an  electromotive  force  of  1.434  volts. 

Ohms  law,  equation  (3)  states  that  the  amperes  flow- 
ing in  any  circuit  are  equal  to  the  volts  divided  by  the 
ohms.  This  is  the  most  important  electrical  law  for  by 
it  when  any  two  of  the  quantities  involved  are  known 
the  third  may  be  determined.  For  instance  if  a  conduc- 
tor of  10  ohms  resistance  has  flowing  through  it  3  am- 
peres, the  voltage  measured  between  its  ends  must  be 
10  X  3  =  30  volts.  Again  if  18  volts  are  applied  to  the 
ends  of  a  conductor  of  9  ohms  resistance  the  current  will 
be  18  -H  9  =  2  amperes.  Finally  if  o.  i  of  an  ampere  is 
desired  from  a  source  whose  electromotive  force  is  two 
volts,  the  total  resistance  of  the  circuit  must  be  20  ohms 

for  as  E  =  C  R.      R  =  -—  and  2  -^  o.  i  =20  ohms. 

We  are  now  in  a  position  to  consider  the  relations  be- 
tween electrical  energy,  heat  and  power. 

Joule  has  shown  by  experiment  that  the  relation  be- 
tween the  heat  developed  in  a  circuit  and  the  current 
flowing  through  it  is  expressed  by  the  equation 

V  =  aRtXo.24  (4) 

When 

C  is  the  current  in  amperes 

R  is  the  resistance  in  ohms 

t    is  the  time  in  seconds  that  the  current  has  passed 

V  is  the  heat  in  calories. 


ELECTRIC   CALCULATIONS.  147 

Consequently  2  amperes  passing  through  5  ohms  for 
10  seconds  develop 

4  X  5  X  10  X  0.24  =  48  calories. 

The  constant  0.24  can  be  eliminated  from  equation 
(4)  if  instead  of  expressing  the  heat  developed  in  calo- 
ries we  measure  it  m  joules  :  for  the  jouhy  the  electrical 
unit  of  heat,  is  0.24  of  a  calorie.^ 

So  if  we  designate  the  heat  produced  in  joules  by  W 
the  expression  becomes 

W  =  C^Rt  (5) 

and  the  heat  developed  by  2  amperes  passing  through  5 
ohms  for  10  seconds  is  4  X  5  X  10  =  200  joules. 

Equation  (5)  may  be  written  W  =  C  t  X  C  R.  But 
we  have  already  seen  from  equation  (i)  that  Q  =  C  t 
and  from  Ohms  law  that  E  =  C  R ;  substituting  these 
values  we  get : 

W  =  Q  E  (6) 

This  equation  (6)  shows  that  the  heat,  measured  in 
joules,  generated  in  a  conductor  is  equal  to  the  number 
of  coulombs  that  have  passed  multiplied  by  the  voltage 
between  the  ends  of  that  conductor. 

For  example,  if  1000  coulombs  (10  amperes  for  one 
minute  and  forty  seconds)  have  passed  through  a  con- 
ductor, the  voltage  between  whose  ends  is  10 ;  there  have 
been  generated  in  that  time  1000  joules,  equal  to  238.  if 
calories. 

Since  heat  is  a  form  of  energy,  the  energy  applied 
to  the  conductor  expressed  in  joules  is  Q  E.     So  that  the 

*  The  latest  determinations  give  one  calorie  equal  to  4.2  joules  which  makes   • 
this  constant  0.2381. 

f  From  this  point  one  calorie  will  be  taken  as  4.2  joules. 


148  ELECTRIC    CALCULATIONS. 

rate  of  work  or  power  expressed  in  joules  per  second  is 
^ —  or  since  ^  =  C  from  equation  (  i )  ;  the  power  ex- 
pressed in  joules  per  second  equals  C  E. 

A  joule  per  second  is  called  a  watt  so  if  we  desig- 
nate the  power  expressed  in  watts  by  P  we  have 

P  =  C  E  (7) 

or  the  number  of  watts  in  any  electrical  circuit  is  equal 
to  the  product  of  the  amperes  and  the  volts.  A  circuit 
in  which  4.2  amperes  are  flowing  at  a  pressure  of  100 
volts  is  performing  work  at  the  rate  of  420  watts,  cor- 
responding to  100  calories,  per  second. 

The  relations  between  the  electrical  and  mechanical 
units  of  power  are  as  follows  : — 

Rowland  has  shown  that  it  requires  780  foot  pounds 
of  energy  to  heat  one  pound  of  water  one  degree 
Fahrenheit,  or  3.1  foot-pounds  to  heat  one  gram  of 
water  one  degree  centigrade,  one  calorie.  Since  one 
calorie  is  equal  to  4.2  joules,  one  foot-pound  is  equal  to 
1.356  joules  or  one  joule  equal  to  0.737  foot-pound. 

550  foot-pounds  therefore  equal  746  joules,  and  since 
550  foot-pounds  per  second  are  one  horse-power  and  a 
joule  per  second  is  a  watt,  746  watts  equal  one  horse- 
power. 

The  practical  unit  for  measuring  large  quantities  of 
electrical  power  is  the  kilowatt,  one  thousand  watts. 
This  is  equal  therefore  to  y2>7  foot-pounds  per  second  or 
approximately  i^  horse-power. 

ELECTROLYTIC  CALCULATIONS. 

When  an  electric  current  passes  between  platinum 
plates  immersed  in  water  acidified  with  sulphuric  acid, 


ELECTROLYTIC  CALCULATIONS.  149 

water  is  decomposed,  oxygen  is  liberated  at  the  platinum 
anode  and  hydrogen  at  the  platinum  cathode :  anode 
and  cathode  are  the  names  given  to  the  electrodes  by 
which  the  current  enters  (anode)  and  leaves  (cathode) 
the  liquid.  There  is  no  liberation  of  hydrogen  or  oxy- 
gen except  at  the  contact  of  the  liquid  with  the  elec- 
trodes. This  chemical  decomposition  effected  by  the 
electric  current  is  called  electrolysis  and  the  bath  or  ma- 
terial which  undergoes  decomposition  the  electrolyte. 

Electrolytes  may  be  solids,  liquids,  or  as  recent  ex- 
periments have  shown,  gases.  Silver  iodide  is  an  exam- 
ple of  a  solid  electrolyte,  while  as  liquid  electrolytes  we 
have  solutions  of  mineral  salts  and  acids  as  well  as  many 
fused  salts.  An  electrolyte  must  necessarily  be  a  con- 
ductor of  electricity  and  in  the  case  of  liquids  be  ionized. 
Liquids  such  as  pure  water,  ether,  and  carbon  bisulphide 
which  are  not  perceptibly  ionized  are  not  considered 
electrolytes. 

The  constituents  into  which  the  electrolyte  is  split 
up  by  the  current  are  called  ions ;  that  which  is  liberated 
at  the  anode  is  the  anion,  that  which  is  liberated  at  the 
cathode  the  cathion.  With  very  few  exceptions  an  ele- 
ment or  radical  is  always  liberated  at  the  same  elect- 
rode. Ostwald*  gives  the  following  classification  of 
ions — 

CATHIONS. 

Monovalent :  H  (in  acids)  K.  Na.  Li.  Cs.  Rb.  Tl. 
Ag.  NH4.  NH3R  to  NR4  (R  being  an  organic  radical) 
Cu  (cuprous),  Hg  (mercurous). 

Divalent :  Ca.  Sr.  Ba.  Mg.  Fe  (ferrous),  Cu  (cupric)., 
Pb.  Hg  (mercuric),  Co.  Ni.  Zn.  Cd. 

*  Scientific  Foundations  of  Analytical  Chemistry  p.  53. 


150  ELECTROLYTIC    CALCULATIONS. 

Trivalent :  Ah  Bi.  Sb.  Fe  (ferric)  and  some  of  the 
rarer  earth  metals. 

Tetravalent  Sn  (?).  Zr. 

ANIONS. 

Monovalent  :  OH  (in  bases)  F.  CI.  Br.  I.  NO3.  CIO3. 
CIO4.  Br04.  Mn04  (in  permanganates)  and  the  anions 
of  all  other  monobasic  acids  the  (acid  molecule  minus 
one  hydrogen  which  goes  to  the  cathode). 

Divalent:  S.  Se.  Te  (?).  SO4.  Se04.  Mn04  (in  man- 
ganates),  and  the  anions  of  dibasic  acids. 

Tri-to  hexavalent:  The  anions  of  the  tri-to  hexaval- 
ent  acids.  Elementary  anions  with  a  valency  of  more 
than  two  are  not  known. 

The  quantity  of  the  ions  deposited  by  the  passage  of 
a  current  through  an  electrolyte  was  shown  by  Faraday 
to  be  very  simply  related  to  the  quantity  of  electricity 
which  passes. 

Faraday's  first  law  of  electrolysis  states :  that  the 
quantity  of  an  electrolyte  decomposed  by  the  passage  of 
a  current  of  electricity  is  directly  proportional  to  the 
quantity  of  electricity  which  passes  through  it.  So  as 
long  as  the  quantity  of  electricity  remains  the  same 
it  is  immaterial,  so  far  as  quantity  deposited  is  con- 
cerned, whether  the  electricity  passes  as  a  very  intense 
current  for  a  short  time  or  as  a  very  weak  current  for  a 
long  time. 

Faraday's  second  law  of  electrolysis  states :  that,  if 
the  same  quantity  of  electricity  passes  through  different 
electrolytes,  the  weights  of  the  different  ions  liberated  will 
be  proportional  to  the  chemical  equivalents  of  the  ions. 
Thus  if  the  same  current  passes  through  a  series  of 
electrolytes   from   which  it   liberates   as   ions   oxygen, 


ELECTROLYTIC   CALCULATIONS.  151 

hydrogen,  silver,  and  chlorine,  then  for  every  8  grams  of 
oxygen  evolved  there  will  be  liberated  1.008  grams  of 
hydrogen  107.93  grams  of  silver  and  35.45  grams  of 
chlorine. 

The  electrochemical  equivalent  of  a  substance  is  the 
weight  in  grams  deposited  by  the  passage  of  one 
coulomb  of  electricity.  Using  this,  Faraday's  laws  can 
be  comprised  in  the  statement :  that  the  number  of 
grams  of  an  ion  deposited  during  the  passage  of  a  cur- 
rent through  an  electrolyte  is  equal  to  the  number  of 
coulombs  that  have  passed  multiplied  by  the  electro- 
chemical equivalent  of  the  ion.  Since  one  ampere  per 
second,  from  the  definition  of  unit  of  current,  deposits 
0.001118  gram  of  silver  and  the  chemical  equivalent  of 
silver  is  107.93,  the  weight  of  any  other  element  liber- 
ated by  one  ampere  per  second  will  be  found  by  the 
proportion. 

0.001 1 18  :  X  :  :  107.93  :  A. 

in  which  A  is  the  chemical  equivalent  of  the  element,  or 

A  (0.001118) 
X  =  — ^^ ^ 

107.93. 

Therefore  since  ^'^^^^^^   =    0.00001036,    the   total 
107.93 

weight  deposited  must  be : 

w  =  0.00001036  A  C  t.  (8) 

In  which — 

w  is  the  weight  in  grams  deposited. 

A   is    the   chemical  equivalent,   that  is  the  atomic 
weight  divided  by  the  valency  of  the  deposited  ion. 

C  is  the  current  in  amperes  and 

t   is  the  time  in  seconds  that  the  current  flows. 


152  ELECTROLYTIC   CALCULATIONS. 

Expressing  the  time  in  hours  and  denoting  it  by  N, 
equation  (8)  becomes : 

w  =  0.0373  A  C  N.  (9) 

C  N  being  ampere-hours. 

By  transposing  this  equation,  we  get  an  expression 
for  the  amount  of  current  required  to  deposit  a  given 
weight  in  a  given  time. 

f^  26.8     W  /  V 

and  since  from  (7)  the  power  in  watts  (P)  is  equal  to 
the  product  of  the  amperes  times  the  electromotive  force 
(C  E)  we  get 

■n        26.8  w  E  ^     V 

An  expression  for  the  power  in  watts  necessary  to  de- 
posit w  grams  of  a  metal  whose  chemical  equivalent  is 
A  in  N  hours  when  an  electromotive  force  E  is  used. 

To  illustrate  the  application  of  these  formulae : 

The  amount  of  oxygen  liberated  during  the  decom- 
position of  acidulated  water  by  a  current  of  2  amperes 
lasting  15  minutes  is  given  by  equation  (8) 

w  =  0.00001036  X  8  X  2  X  900  =  1.492  grams. 

The  amount  of  copper  deposited  in  3  hours  by  a  cur- 
rent of  25  amperes  from  a  solution  of  copper  sulphate  is 
found  by  equation  (9),  to  be 

w  =  0.0373  X  31-8  X  25  X  3  =  8-895  grams. 

To  determine  the  number  of  amperes  that  are  neces- 
sary to  deposit  500  grams  of  nickel  in  5  hours,  we  em- 
ploy formula  (10)  and  find 

f^      26.8  X  soo 

C  = -^ =  91.3  amperes. 

29.35  X  5 

The  power  necessary  to  deposit  48.8  grams  of  zinc 


COUNTER-ELECTROMOTIVE   FORCE.  153 

in  four  hours  with  a  voltage  of  four  is  found  by  substitu- 
in  equation  (i  i),  to  be  40  watts. 

r,       26.8  X  48.8  X  4 

P  =  1 !t  =  40  watts. 

32.7  X  4 

The  fact  that  elements  which  form  two  series  of  salts 
have  different  electro-chemical  equivalents  according  to 
the  solution  from  which  they  are  deposited,  must  be  borne 
in  mind  when  these  calculations  are  made.  For  example 
copper  forms  two  chlorides  CugCls  and  CuCL.  In  the 
former  its  chemical  equivalent  is  twice  its  value  in  the 
latter :  hence  the  actual  weight  of  copper  deposited  is 
twice  as  great  when  a  solution  of  cuprous  chloride  is 
electrolysed  as  with  a  solution  of  cupric  chloride;  the 
current  and  time  being  the  same. 

COUNTER-ELECTROMOTIVE    FORCE, 

When  a  molecule  is  formed  by  the  combination  of  one 
or  more  elements,  a  definite  amount  of  heat  is  evolved  or 
absorbed.  In  most  stable  compounds  heat  is  evolved  : 
to  decompose  such  a  molecule  energy  must  be  supplied 
which  is  equivalent  to  this  amount  of  heat.  This  energy 
can  be  supplied  by  an  electric  current.  To  separate 
metallic  zinc  from  zinc  sulphate  requires  energy  to  be 
given  to  the  solution  equivalent  to  the  quantity  of  heat 
produced  by  the  formation  of  zinc  sulphate.  Conse- 
quently when  an  electric  current  passes  through  an 
electrolyte,  part  of  the  electrical  energy  furnishes  the 
energy  necessary  to  overcome  the  chemical  forces  and 
only  the  remainder  produces  heat  according  to  Joule's 
law.  The  electrolytic  cell  has  therefore  a  counter-elec- 
motive  force  resisting  the  passage  of  or  opposed  to  the 
current  and  the  electromotive  force  producing  that  cur- 
rent. 


154  COUNTER-ELECTROMOTIVE   FORCE. 

Calling  this  counter-electromotive  force  e,  the  energy 
supplied  must  be   a  certain  number  of  joules  such  that 
the  electrical  energy  =  Qe  =  Cet ;  depending  upon  the 
number  of  coulombs  that  have  passed. 
(Compare  equation  6) 

Or  expressed  in  calories, 

Electrical  Energy  =  —  Cet 

Now  the  energy  of  combination  of  various  compounds 
may  be  measured  by  the  heat  which  is  liberated  when 
the  combination  takes  place,  that  is 

Chemical  Energy  =  w  h, 
when  w  is  the  weight  of  the  element  in  grams  and  h  is 
the  number  of  calories  produced  when  one  gram  of  this 
metal  combines  with  the  other  constituents  of  the  com- 
pound. 

Now  from  Faraday's  law  of  electrolysis  as  given  by 
equation  (8)  we  have  that  w,  the  weight  of  metal  de- 
posited =  0.00001036  A  C  t,  substituting  this  value  for 
w,  we  have 

Chemical  Energy  =  0.00001036  ACth         (12) 

This  chemical  energy  must  equal  the  electrical  energy 
supplied  to  the  cell  available  for  electrolysis,  that  is  the 
electrical  energy  not  producing  heat. 
Therefore 

Cet  =  0.00001036  ACth 

which  by  cancelling  Ct  and  reducing  gives 

e  =  0.0000435  Ah  (13) 

So  the  counter  electromotive  force  is  equal  to  the  heat 
of  combination  of  one  gram  in  calories  times  the  chemi- 
cal equivalent  times  a  constant,  0.0000435. 

Instead  of  stating  the  heat  of  combination  in  terms  of 


COUNTER-ELECTROMOTIVE    FORCE.  155 

one  gram  of  the  metal,  the  more  convenient  method  for 
electrolytic  calculations  is  to  state  it  in  terms  of  the 
chemical  equivalent  in  grams.  For  example,  the  chemi- 
cal equivalent  of  zinc  is  32.7  so  that  the  heat  of  combi- 
nation of  32.7  grams  of  zinc  with  other  substances  is  the 
value  to  be  used.  When  the  values  are  given  in  terms 
of  the  molecular  weight  in  grams  these  must  be  reduced 
to  the  heats  of  combination  per  equivalent  by  dividing 
by  the  valency  or  multiple  of  the  valency  of  the  element. 
The  heat  of  formation  of  Al2(S04)3  is  given  by  Thomp- 
son as  150630  calories:  this  means  the  combination  of 
54.2  grams  of  aluminium  with  sulphur  and  oxygen  to  give 
342.38,  the  molecular  weight  in  grams  of  aluminium  sul- 
phate. As  the  equivalent  of  aluminium  is  9.033  we  must 
divide  by  6  (twice  the  valency  as  there  are  two  atoms)  to 
obtain  the  heat  per  equivalent,  25315  calories. 

Calling  the  heat  per  chemical  equivalent  in  grams 
H  and  remembering  that  A  is  the  chemical  equivalent 
and  h  is  the  heat  evolved  by  the  combination  of  one  gram, 

H  =  Ah, 
substituting  this  value  in  equation  (13)  we  get 
e  =  0.0000435  H  and 

since   0.0000435    equals  — ^-—     this    equation    reduces 

without  appreciable  error  to  the  very  simplest  form  of 
Thompson's  law. 

e  =  -^  (H) 

23000 

A  definite  case  may  make  this  deduction  clearer. 
When  31.8  grams  of  copper,  (the  weight  of  the  chemical 
equivalent  in  grams)  combines  to  form  copper  sulphate, 
27980  calories  are  developed.  Now  if  no  energy  is  lost 
as  heat  the  electrical  energy  necessary  to  deposit  31.8 


156  COUNTER-ELECTROMOTIVE   FORCE. 

grams  of  copper  from  copper  sulphate  must  equal  this. 
This  electrical  energy  is  the  product  of  e,  the  counter- 
electromotive  force,  and  Q,  the  number  of  coulombs  ne- 
cessary to  deposit  31.8  grams  of  copper.  We  find  the 
number  of  coulombs  by  substituting  in  equation  (8) 
giving 

31.8  =  0.00001036  X  3^'^XQ 

or  Q  = =  96525  coulombs. 

0.00001036 

Now  as  one   calorie  equals  4.2  electrical  heat  units 

WG  have 

27980  X  4-2  =e  X  96525 

27980  X  4.2  .     ,,  27080 

or  e  =  —^-^ — — — - —  or  practically  e  =  -^ —  =  1.22 
96525  23000 

IT 

volts  agreeing  with  formula  (14)  e  =  where  H  is 

23000 

the  heat  produced  by  the  combination  of  the  chemical 

equivalent  in  grams  expressed  in  calories. 

This  exceedingly  simple  formula"^  allows  the  elec- 
tromotive force  of  any  chemical  reaction,  whether  it  be 
direct  as  in  a  voltaic  cell,  or  counter  as  in  an  electrolytic 
cell,  to  be  easily  calculated  from  the  heats  of  combi- 
nation. The  calculated  values  agree  in  many  cases  very 
closely  with  those  found  by  experiment.  The  discrep- 
ancies probably  are  due  to  the  fact  that  all  the  energy 
of  combination  in  the  case  of  a  voltaic  cell,  for  instance, 
may  not  be  converted  into  electrical  energy,  as  some 
may  be  converted  into  heat  which  appears  in  the  cell. 

To  calculate  the  voltage  of  a  Daniel's  cell :  zinc  is  dis- 
solved at  one  pole  to  form  zinc  sulphate,  the  electromotive 

*  It  is  also  applied  to  fused  electrolytes.  See  Langley,  J.  Am.  Chem.  Soc, 
vol.  16,  p.  52,  1894. 


COUNTER-ELECTROMOTIVE   FORCE.  157 

force    is    (14)   5.^^  =  2.31   volts  :   at  the  other   pole 
23000  ^ 

copper  is  deposited  which  sets  up  a  counter-motive  force 
found  in  the  same  way  to  be  1.22  volts.  So  that  the 
available  electromotive  force  of  the  cell  (neglecting  the 
slight  electromotive  force  where  the  two  solutions  touch) 
is  equal  to  2.31  —  1.22  or  1.09  volts.  The  electromotive 
force  determined  experimentally  is  about  1.08  volts. 
The  table  at  the  back  of  the  book  gives  some  of  the  most 
important  values  of  H  to  be  substituted  in  equation  14. 
If  other  tables  are  used,  the  different  methods  of  express- 
ing calorific  power  must  be  considered  also  in  some 
cases  the  heat  of  solution  and  that  of  any  chemical  com- 
binations which  take  place  in  the  cell  and  so  give  energy 
which  aids  the  current.  For  example  when  copper  sul- 
phate is  decomposed  in  an  aqueous  solution  we  have  : 
Cathodes— oCu  |  SO4— H.SO4— °  Ho  |  O  o— =>Anode. 

We  have  in  reality  two  decompositions  and  one  com- 
bination so  the  electrical  energy  to  be  supplied  is  that 
equivalent  to  the  heat  of  combination  of  CUSO4  +  its 
heat  of  solution  in  water  +  the  heat  of  combination  of 
H2O  —  (the  heat  of  combination  of  sulphuric  acid  +  the 
heat  developed  on  diluting  with  water). 

When  the  heats  of  combination  are  taken  from  Ost- 
wald's  tables  H  is  obtained  as  follows  for  copper  sulph- 
ate and  nickel  sulphate 

CUSO4  1826K     NiS04aq. 

Heat  of  solution       1 58K     HoO 
HoO  684K 


2668K     H2S04aq. 
H2S04aq.  2109K 

559K 
As    these  values  are   based    on    molecular    weights    in 


158  OHM'S   LAW   FOR   ELECTROLYSIS. 

grams  to  be  converted  to  equivalent  weights  they  must 
each  be  divided  by  two,  since  copper  and  nickel  are 
divalent  ions ;  and  since  K  =  loo  calories  we  get, 

-Izil  =1.22  volts  for  CUSO4 

230 

and  4M^  =  1.89  volts  for  NiSO^ 

230 

The  method  of  calculation  just  described  is  not  in 
accord  with  the  present  views  on  the  ionization  of  elec- 
trolytes. In  order  to  reconcile  the  two,  Nernst's  solution 
tension  theory  must  be  studied.  This  theory,  when  the 
necessary  data  are  at  hand,  leads  to  very  similar  results 
for  the  counter  electromotive  force  and  is  more  accurate 
as  it  takes  into  consideration  variations  in  dilution. 

The  plan  given  here  is  retained  as  being  of  practical 
use  on  account  of  the  great  number  of  heats  of  combina- 
tion which  have  been  determined  with  accuracy  and  the 
simplicity  of  the  calculation. 

OHM'S  LAW  FOR  ELECTROLYSIS. 

In  a  circuit  where  electrolytic  work  is  being  done 
the  quantities  of  electricity  are  not  related  in  the  form 
of  Ohm's  law  already  given  (3)  as  the  counter-electro- 
motive force  must  be  considered.  The  total  electrical 
energy  supplied  to  a  circuit  is  given  by  equation  (6)  as 
equal  to  QE  or  CEt. 

Similarly  the  chemical  work  done  is  Qe  or  Get. 

The  heat  generated  is,  from  equation  (5),  C^Rt. 
Now  from  the  principle  of  conservation  of  energy  we 
know  that  the  electrical  energy  supplied  must  equal  the 
sum  of  the  heat  produced  and  the  chemical  work  done. 
Hence  CEt  =  C»  Rt  +  Get 

Cancelling  G  and  t  we  obtain 

E  =  GR  +  e.  (15) 


OHM'S   LAW   FOR   ELECTROLYSIS.  169 

Where  E  is  the  electromotive  force  required  to  produce 
the  current  C,e  the  counter-electromotive  force  of  the 
electrolytic  cell  and  R  the  total  ohmic  resistance  in  the 
circuit. 

R  may  be  separated  into  its  constituent  parts  by 
calling  Rg  the  resistance  of  the  generator,  Re  the  resist- 
ance of  the  conductors  including  any  extra  resistance 
inserted  to  regulate  the  current  and  Re  the  resistance 
of  the  electrolytic  cell :  then  equation  (i  5)  takes  the  form 

E  =  C  (Rg  +  Re  +  Re)  +  e      (16) 
and  solving  for  C  we  have 

C  =   ^   -    ^ (17) 

Rg  +  Re  +  Re  ^  '^ 

The  effective  electromotive  force  in  the  circuit  is 
lower  than  the  electromotive  force  of  the  generator  by 
the  counter-electromotive  force  of  the  electrolytic  cell, 
so  the  current  is  equal  to  the  difference  of  these  electro- 
motive forces  divided  by  the  total  resistance.  For  any 
value  of  E  smaller  than  e,  no  current  will  flow  and  no 
metal  be  deposited  ;  so  this  counter-electromotive  force 
is  also  the  value  which  the  applied  electromotive  force 
must  exceed  in  order  to  produce  decomposition. 

This  fact  permits  the  deposition  of  a  single  metal 
from  solutions  containing  several.  For  instance  the  heat 
of  combination  per  equivalent  of  nickel  sulphate  in 
water  is   43475   calories   which   means   that  it   requires 

43475  ^  J  gc|  volts  to  cause  its  decomposition  and 
23000 

deposit  nickel :  to  deposit  copper  from  copper  sulphate 
requires  1.22  volts.  Therefore  if  we  apply  a  voltage 
above  1.22  and  below  1.89  to  a  cell  containing  a  mix- 
ture of  copper  and  nickel  sulphate  in  solution  only 
copper  will  be  deposited.     Given  a  solution  containing 


160 


OHM'S    LAW   FOR   ELECTROLYSIS. 


silver  and  copper  as  nitrates.  What  are  the  limits  of 
voltage  to  deposit  silver  and  not  copper  ?  The  heats  of 
combination  per  equivalent  of  these  salts  in  aqueous 
solution  are  respectively  8390  and  26205  calories,  div- 
iding by  23000  we  get  the  counter-electromotive  force 
as  0.365  volt  for  silver  nitrate  and  1.14  volts  for  copper 
nitrate.  So  any  current  with  a  voltage  above  0.365  and 
below  1.14  will  deposit  only  silver  from  this  solution. 

When  all  of  the  first  metal  is  deposited  the  current 
will  fall  to  zero  unless  the  voltage  is  raised  above  the 
counter-electromotive  force  of  the  next  higher  metal  in 
solution. 

A  convenient  method  of  regulating  the  voltage  is  shown  by 
the  diagram. 

A  and  B  are  the  poles  of  the  source  of  current.     The  circuit 

is  closed  through  a  conductor  CD  whose  resistance  is  sufficient 

to  cut  down  the  current.     The  resistance  of  this  conductor  must 

of  course  depend  on  the  source  of  current,  which  may  be  either 

A  B  primary  batteries,  second- 

-^  ^ ^       ary   batteries,  dynamo   or 

Q  lighting  circuit.  The  elec- 
trolytic cell  is  connected  so 
that  one  pole  E  connects 
with  C  and  the  other  pole 
F  connects  with  the  mov- 
able point  or  slider  G  so 
that  any  point  of  the  con- 
ductor CD  can  be  selected 
and  any  voltage  from  zero 
(when  G  is  at  C)  to  the  full 
voltage  of  the  source  (when  G  is  at  D).  By  having  a  voltmeter 
V  connected  across  the  cell  and  starting  with  G  at  C  the 
proper  voltage  can  be  applied  by  pushing  G  towards  D  until  the 
voltmeter  indicates  the  value  desired. 

The  relation  existing  between  the  voltage  measured 


OHM'S   LAW   FOR    ELECTROLYSIS.  161 

across  an  electrolytic  cell  by  a  voltmeter  and  the  current 
passing  through  the  cell  is  given  by  equation  (i6);  when 
Rg  and  Re  are  made  zero,  this  becomes 

E  =  CRc  +  e  (i8) 

If  Re  be  negligible,  then  unless  C  is  very  great  E  = 
e,  or  the  voltage  across  the  cell  is  practically  a  constant, 
equal  to  the  counter-electromotive  force  of  the  cell  and 
independent  of  the  current  sent  through  it. 

If  either  Re  or  C  is  so  great  that  e  may  be  neglected 
then  the  voltage  is  proportional  to  the  current  flowing 
as  in  the  case  of  simple  conductor. 

In  most  cases,  however,  neither  can  be  neglected 
and  the  voltage  across  the  cell  depends  not  only  on  the 
counter-electromotive  force  but  also  on  the  resistance 
of,  and  the  current  flowing  through  the  electrolytic  cell. 
Let  us  consider  an  electrolytic  cell  containing  a  satura- 
ted solution  of  zinc  sulphate  with  platinum  electrodes  5 
centimeters  apart  whose  opposite  submerged  surfaces 
are  15  square  centimeters,  since  the  specific  resist- 
ance is  2>2)'7  ohms  Re  in  this  case  will  be  (equation  (2)), 

2>2>'7  X   —  =   11.23  ohms. 

The  counter-electromotive  force  of  zinc  sulphate  is 
2.31  volts  (see  tables)  if  the  current  passing  is  one-tenth 
of  an  ampere,  the  voltage  across  the  cell  will  be  found 
by  equation  (18) 

E  =  0.1  X   11.23  +   2.31    =  3.43  volts. 

If  the  current  is  raised  to  two-tenths  of  ampere  the 
voltage  will  not  be  double,  6.86,  but  only  4.56  volts. 
For 

E  =  0.2  X   11.23  +  2.31   =  4.56  volts. 

If  the  cell  were  so  arranged  that  its  internal  resist- 


162  EFFECT     OF   DISSOLVING  ANODES. 

ance  were  only  one-tenth  of  an  ohm  for  small  currents 
the  voltage  across  its  poles  would  be  practically  constant : 
for  0.05  of  an  ampere,  2.315  volts ;  for  o.  10  of  an  ampere  ; 
2.32  volts. 

Silver  nitrate  in  water  requires  only  0.365  of  a  volt  to 
effect  decomposition.     Therefore,  for  such  a  cell,  having 
a   resistance   of    100    ohms,    with    Y4   ampere  passing 
through  it,  the  voltage  across  the  poles  will  be 
E  =  V4  X   100  +  0.365  =    25.365. 

When  the  current  is  doubled  this  becomes 

E  =  V2  X   100  +  0.365   =   50.365. 
or  practically  double. 

EFFECT    OF    DISSOLVING    ANODES. 

In  plating  processes  and  the  electrolytic  refining  of 
metals^  the  anode  of  the  bath  is  made  of  the  same  metal 
as  that  to  be  deposited,  so  that  as  the  current  passes  the 
anode  dissolves  at  the  same  rate  at  which  the  metal  is 
deposited  on  the  cathode ;  keeping  the  solution  at  the 
same  strength.  A  plating  or  refining  bath,  or  any  cell 
with  a  dissolving  anode  of  the  same  metal  as  in  the 
solution,  has  therefore  no  counter-electromotive  force, 
since  the  electromotive  forces  of  solution  and  decompo- 
sition are  equal  and  opposed  to  each  other. 

Take  the  familiar  example  of  a  cell  containing  copper 
sulphate,  the  anode  of  platinum,  the  cathode  of  copper; 
this  cell  has  a  counter-electromotive  force  of  1.22  volts. 
When  we  substitute  a  copper  anode  for  the  platinum, 
no  sulphuric  acid  is  formed  but  the  copper  forms  copper 
sulphate  with  the  SO4  and  the  anode  dissolves  at  the 
same  rate  as  the  copper  is  deposited.  The  formation  of 
copper  sulphate  ordinarily  generates  27,980  calories  per 
equivalent  in  grams,  but  here  the  energy  appears  as 


CURRENT   DENSITY.  163 

electricity  and  the  electromotive  force  produced  is  ex- 
actly equal  to  that  necessary  to  decompose  copper  sul- 
phate. A  bath  or  cell  of  this  kind  only  introduces  a 
certain  ohmic  resistance  into  the  circuit  (its  e  being 
zero)  and  for  it  formulae  15,  16,  17,  18,  reduce  to  the 
simplest  form  of  Ohm's  law.  (3) 

CURRENT    DENSITY. 

The  character  of  the  deposit  depends  largely  on  the 
strength  of  current  used,  which  can  only  be  determined 
by  experiment.  When  this  has  been  found,  the  current 
is  often  described  as  having  a  certain  density,  that  is  a 
certain  number  of  amperes  per  unit  of  cathode  area. 

In  analytical  work  current  density  is  expressed  in 
amperes  per  100  square  centimeters  of  cathode  area; 
that  is  area  on  which  metal  is  deposited.  N.D.joo  is  the 
abbreviation  used:  so  N.D.joo  0.25  means  a  quarter  of 
an  ampere  for  each  160  square  centimeters  of  the  sub- 
merged area  of  the  cathode  on  which  metal  is  deposited. 
As  it  is  so  important  in  this  work  to  have  a  firm  adher- 
ent  deposit  the  current  is  most  conveniently  described 
by  giving  the  voltage  and  current  density.  In  some  of 
the  older  books  on  the  subject  the  current  is  given  in 
cubic  centimeters  of  mixed  gases  per  minute,  in  order  to 
reduce  this  to  current  density  it  is  necessary  to  know 
that  one  ampere  equals  10.436  c.c.  of  oxyhydrogen  gas 
per  minute  measured  at  o°C  and  760  millimeters 
pressure. 


EXAMPLES, 

WITH    SOLUTION    OF    EACH. 


1.  How  many  amperes  are  flowing  in  a  circuit  which 
in  the  course  of  one  hour  and  1 5  minutes  has  deposited 
30.186  grams  of  silver  ?  Ans,     6  amperes. 

Solution.      I  hour  and  15  minutes  =  4500  seconds  : 

one  ampere  by   definition  deposits  0.001 118    gram  of 

silver  per  second.     Therefore    in  4500  seconds  it  will 

deposit  0.001118  X  4500,  or  5.031   grams.     Number  of 

amperes  flowing  equals  30.186  divided  by  5.031  or  6. 

2.  If  a  current  of  15  amperes  flows  for  one  hour  and 
20  minutes  how  many  coulombs  will  pass  ?  How  many 
ampere-hours  will  that  give  ? 

Ans,     72000  coulombs,  20  ampere-hours. 

Solution.     Substitute  in  equation  (i) 

Q  =  Ct  =  15  X  4800  =  72000  coulombs:  since  i  hr. 
and  20  minutes  =  4800  seconds.  72000  coulombs  -4- 
3600  =  20  ampere-hours. 

3.  If  the  specific  resistance  of  copper  is  0.000001629, 
what  is  the  resistance  of  a  wire  200  meters  long  having  a 
cross-section  of  2  square  millimeters  ? 

Ans.      1.629  ohms, 

Solution.     Substitute  in  equation  (2)  R  =  p  — 

a 

p  =  0.000001629,  1  =  20000  centimeters,  a  =  0.02  square 

centimeter. 

T»  ^       20000  ^ 

R  =  0.000001629 =  1.629. 

0.02 

( 164 ) 


EXAMPLES.  165 

4.  If  the  specific  resistance  of  German-silver  is 
0.0000209,  what  is  the  resistance  of  150  meters  of  wire 
having  a  cross-section  of  25  square  millimeters. 

Ans,      1.254  ohms. 

Solution.     Substitute    in    equation  (2)  R  =  p 

a 

P  =  0.0000209,  1  =  15000  centimeters,  a  =  0.25  square 

centimeter 

R  =  0.0000209  —5 =  0.0000209  X  60000  =  1.254 

0.25 

5.  What  is  the  resistance  of  an  electric  light  carbon 
1 2  inches  long  by  ^/^  inch  in  diameter  when  its  specific 
resistance  is  0.07.  It  will  be  sufficient  to  call  2  %  centi- 
meters one  inch.  Ans.     7  ohms. 

Solution.     Substitute  in  equation  (2)  R  =  p  ^ — 

a 

p  =  0.07,  1  =  12  X  2.5  =  30  centimeters,  a  =  ;rr^  = 

Tt  (y^  X  2.5)2  =  0.3  very  closely.  R  =  0.07  ^    =    7  ohms 

*  o 
practically. 

6.  What  is  the  resistance  of  a  saturated  solution  of 
sulphate  of  copper  placed  in  a  tube  10  cm.  long  and 
having  a  cross-section  of  2  square  centimeters,  when  its 
specific  resistance  is  29.3.  Ans.      146.5  ohms. 

Solution.     Substitute  in  equation  (2)  R  =  p  — 

p  =  29.3,  1  =  10,  a  =  2.    R  =  29.3  i^  =  146.5 

7.  What  is  the  total  resistance  in  a  circuit  composed 
of  a  voltaic  cell,  a  copper  wire,  an  iron  wire  and  a  piece 
of  German-silver  all  connected  in  series  when  the  separ- 
ate resistances  are  respectively  3,  2,    4,  6  ? 

Ans.     15  ohms. 


166  EXAMPLES. 

Solution.  Total  resistance  of  resistances  connected 
in  series  equals  their  sum. 

8.  What  is  the  resistance  of  eleven  incandescent 
lamps  connected  in  parallel  if  each  has  a  resistance  of 
220  ohms  when  hot.  Ans.     20  ohms. 

Solution.  Resistance  of  equal  resistance  connected 
in  parallel  is  equal  to  resistance  of  one  divided  by  their 
number. 

9.  What  is  the  resistance  of  four  wires  of  respectively 
20,  5,  4,  and  2  ohms  connected  in  parallel. 

Ans.      One  ohm. 
Conductance  of  a  number  of  conductances  connected 

in  parallel  equals  their  sum.  — '    — '  — '  —  equals  .05   + 

20    5     4      2 

.2  +  . 25  +  . 50=  I.     Reciprocal  of  one  is  one. 

10.  If  6  amperes  are  flowing  through  7  ohms,  what  is 
the  voltage  measured  between  the  ends  of  this  resistance? 

Ans.     42  volts. 
Solution.     Ohm's  law,  E  =  CR  =  6  X  7  =  42. 

11.  If  a  voltage  of  17  is  applied  to  34  ohms  how 
many  amperes  will  pass?  Ans,     0.5  ampere. 

Solution.     C  =  —  =  _Z  =  0.5 
^      34 

1 2  What  is  the  resistance  of  a  conductor  that  allows 

%  of  an  ampere  to  flow  when  a  voltage  of  75  is  applied 

to  its  terminals ?  Ans.     100  ohms. 

E         7  ^ 
Solution.     From  Ohm's  Law  R  =  —  =  J-^=  100 

C       0.75 

13.   If  a   wire  of  2    ohms    resistance   is    immersed 

in  100  grams  of  distilled   water  how  long  will  it   take 

^4  of  an  ampere  to  raise  the  temperature  of  the  water 

from  4  to  6  degrees  centigrade,  assuming  no  loss  from 

radiation.  A 71s.      12  minutes  21  seconds. 


EXAMPLES.  167 

Solution.  loo  grams  of  water  raised  2^  =  200  calor- 
ics.    Substitute  in  formula  (4) 

V  -=  C^Rt  X  0.24.  V  =  200,  C  =  7„  R  =  2. 

200  =  -^  X  2  X  t  X  0.24  =  0.27  t.       t  =  741  seconds 
10 

or  12  minutes  21  seconds. 

14.  How  many  joules  would  be  expended  as  heathy 
3  amperes  flowing  through  2  ohms  for  10  minutes. 

Ans.     10,800  joules. 
Solution.     Substitute  in  equation  (5)  W  =  C^Rt. 
C  =  3,  R  =  2,  t  =  600.  W  =  9  X  2  X  600  =  10800. 

15.  An  enclosed  arc  lamp  requires  5  amperes  to 
burn  properly.  When  placed  on  an  118  volt  circuit, 
how  many  watts  are  taken  ?  Ans.     590  watts. 

Solution.     Substitute  in  equation  (7) 
P  =  CE       C  =  5,  E=ii8.     P  =  5Xii8  =  590 

16.  How  many  watts  do  25  incandescent  lamps  take 
connected  in  parallel  from  a  1 10  volt  circuit,  when  each 
lamp  draws  half  an  ampere  ?  How  many  kilowatts  and 
how  many  horse-power  does  this  make  ?  Ans.  1375 
watts,  1.375  kilowatts  and  1.84  horse-power. 

Solution.     25  lamps  at  Y2  ampere  apiece  gives  12.5 
amperes.     Substitute  in  equation  (7)     P  =  CE, 
C  =  12.5,  E  =  no,  P  =  12.5  X  no  =  1375. 
1000  watts  =  I  kilowatt.     1375  watts  =  1.375  kilowatts. 
746  watts  =   I  horse-power.     1375  watts  =1.84  horse- 
power. 

1 7.  At  full  load  a  dynamo  delivers  600  amperes  at 
125  volts  what  is  its  output  in  watts  and  what  would  be 
its  rating  in  kilowatts  and  horse-power. 

Ans.     75000  watts,  75  kilowatts,  100  horse-power. 
Solution.     Substitute  in  equation  (7) 

P  =  CE.     C  =  600,  E  =  125,  P=  75000. 


168  EXAMPLES. 

132. —  =  75  kilowatts,   i   kilowatt  =    i  Vs  horse-power, 

lOOO 

75  X  I  Vs  ~  ^^^  horse-power. 

1 8.  How  much  zinc  will  be  deposited  from  zinc  sul- 
phate by  3  amperes  in  lo  minutes  ?    Ans.  0.6098  gram. 

Solution.     Substitute  in  equation  (8) 
w  =  0.00001036  ACt.  A  =  32.7,   C  =  3,  t  =  600. 
w  =  0.00001036  X  32.7  X  3   X  600  =  0.0003388  X 
1800  =  0.6098  gram. 

19.  How  much  lead  will  be  deposited  by  a  current 
of  Ys  an  ampere  passing  for  90  minutes  through  lead 
chloride.  Ans.     2.894  grams. 

Solution.     Substitute  in  equation  (8) 
w  =  0.00001036  ACt     A  =   103.5,   C  Yo,  t  =  5400. 
w  =  0.00001036  X  103.5  X  Vs  X  5400 
=  0.001072  X  2700  =   2.894  grams. 

20.  How  much  chlorine  would  be  liberated  by  Ys  an 
ampere  passing  through  dilute  hydrochloric  acid  for  13 
hours.  Ans.     8.593  grams. 

Solution.     Substitute  in  equation  (9) 
w  =  0.0373  ACN.     A  =  35.45,  C=  %  N  =  13. 
w  =  0.0373  X  35-45  X  Y2  X  13 
=  1.322  X  6.5  =  8.593  grams. 
21   How  much  tin   will   be  deposited  by  5  amperes 
flowing  6  hours  through  a  solution  ofSnClg? 

Ans.     66.6  grams. 
Solutiun.      Substitute  in  equation  (9) 
w  =  0.0373  ACN.     A  =  59.5,  C  =  5,  N  =  6 
w  =0.0373  X  59-5  X  5  X  6 
w  =  2.220  X  30  =  66.6  grams. 
22.   How  many  amperes  will  it  take  to  deposit  2  J/2 
pounds  of  copper  from  copper  sulphate  in  10  hours? 

Ans.     95.7  amperes. 


EXAMPLES.  169 

Solution.     Substitute  in  equation  (lo) 
C  =     ,  ^^.    w  =  1 136  grams,    i  kilo  =  2.2  lbs.  hence 

2.5  lbs  =  1. 136  kilos  =1136  grams. 

A       .T  Q    XT      T^      n       26.8  X   1 136 

A  =  31.8,  N  =10.     C  = ±-  =  95.7  amperes. 

31.8  X  10  ^ 

23.  What  is  the  minimum  voltage  necessary  to  de- 
compose water  if  the  heat  of  combination  of  one  gram 
of  hydrogen  with  oxygen  is  34180  calories. 

Ans.     1.486  volts. 
Solution.     Substitute  in  equation  (14) 

e=       "      =  34180  =  ,.^86  volts. 

23000  23000 

24.  If  zinc  bromide  requires  1.65  volts  to  decompose 
it,  what  is  the  heat  evolved  by  the  equivalent  weight  of 
zinc  in  grams  combining  with  bromine  ? 

Ans.  37950  calories. 
Solution.     From  equation  (14)  H  =  23000  e  hence 
H  =  23000  X  1.65  =  37950- 

25.  If  one  gram  of  hydrogen  combining  with  chlor- 
ine in  the  presence  of  water  evolves  39315  calories  what 
is  the  voltage  required  for  the  decomposition  of  hydro- 
chloric acid?  Ans,     1.71  volts. 

Solution.     Substitute  in  equation  (14) 

e  =  =  j2z2_^  =  1. 7 1  volts. 

23000       23000 

26.  It  is  desired  to  send  half  an  ampere  through  ten 
ohms  of  copper  sulphate.  What  resistance  must  be  put 
in  series  with  the  cell  if  the  current  is  to  be  drawn  from 
a  circuit  whose  voltage  is  118.  What  will  be  the  values 
of  this  extra  resistance  for  respectively  ^  and  2  am- 
peres ?  Ans,     223.56,  457-12,  and  48.39  ohms. 


170  EXAMPLES. 

Solution.     Neglecting  the  resistance  of  the  generator 
and  leads  equation  (i6)  becomes  for  this  example : 
E=C(Ri+R,)  +  e.  E=  ng,  C  =  %,  Re=  io,e  =  1.22 
118  =  V2(Ri+  10)  +  1.22 

Y3R1  =  118  -  5 — 1.22  =  1 1 1.78      Ri  =  223.56  ohms. 
IfC   =  V,.  118  =  V4(Ri+  10)  +  1.22 

V4R1  =  118-  2.5-  1.22  =  114.28  Ri  =  457. 12  ohms 
IfC  =  2.    118  =  2(Ri+  10)  +  1.22 

2Ri=ii8  —  20  -  1.22  =  96.78.     Ri  =  48.39  ohms. 

27.  Two  batteries,  having  a  negligible  internal  resist- 
ance and  a  voltage  of  two  each,  are  to  furnish  cur- 
rent to  be  sent  through  a  solution  of  zinc  sulphate  of 
2.7  ohms  resistance.  What  extra  resistance  must  be  in- 
serted to  limit  the  current  to  half  an  ampere  ?  How 
much  more  must  be  added  to  cut  the  current  down  to 
Y4  of  an  ampere  ? 

Ans,     0.68  of  an  ohm,  3.38  ohms  additional. 
Solution.     Substitute  in  E  =  C(Ri  +  R^)  +  e 
E  =  4.,  C  =  V2»  Re  =  2.7,  e  =  2.31   (see  tables) 
4  =  V2  (Ri  +  2.7)  +  2.31.     Vg  Ri  =  4  -  1.35  -  2.31. 

Ya  Rj  =  0.34.  Ri  =  0.68  of  an  ohm. 
If  C  =  %  4  =  V4  (Ri+  2.7)  +  2.31.  %  R,  =  4-0.675 
—  2.31  =  1. 01 5.  Ri  =  4.06.  4.06  —  0.68  =  ^.^S  ohms 
additional. 

28.  What  current  will  pass  through  a  tube  of  acidu- 
lated water,  having  a  resistance  of  25  ohms,  between 
platinum  electrodes  when  connected  to  a  118  volt  circuit 
in  series  with  6  lamps  in  parallel  assuming  each  lamp  to 
be  240  ohms  resistance.  What  will  be  the  value  of  the 
current  when  half  the  lamps  are  unscrewed  ? 

Ans.      1.79  and  i.ii  amperes. 
Solution.     Substitute  in  equation  (17) 


EXAMPLES.  171 

E  =  ii8,  e  =  1.486   example   (23)   Rg  =  o,    neglected, 

r^         I  18  -  1.486         I  16.SI4 

C  = —  = ^^— t  =1.70  amperes. 

40+25  65 

For  3  lamps  R,  =  ^  =  80.    C  =  i  iB- 1.486  ^  ij6^4 

^3  80+25  105 

=  1.11  amperes 

29.  What  is  the  voltage  across  an  electrolytic  cell  of 
silver  nitrate  having  a  resistance  of  40  ohms  when  half 
an  ampere  is  passing.  Ans.     20.365  volts. 

Solution.  Substitute  in  equation  (18)  E  =  CR^+e. 
C  =  Y2>  Re  =  40,  e  =  0.365  (see  tables)  E  =  Y2X  40  + 
0.365  =  20.365  volts. 

30.  What  is  the  voltage  across  an  electrolytic  cell  of 

zinc  bromide  of  2  ohms  resistance  when  the  circuit  is 

closed  by  inserting   1Y2  ohms  from   a  storage   battery 

giving  4  volts.  Ans.     3  volts  practically. 

Solution.     First  find  C  from  equation  (17) 
•p  • 
^  ~      .  E  =  4,  e  =  1.65    (example  24)  Rj  =  1.5 


Ri+R, 

Re  =2. 

C  =  .  ^~  ^'^^  =   M^  =067  practically 
1.5+2  3-5 

Now  substitute  in  equation  (18)  E  =  0.67  X  2  +  1.65 
=  1.34  +  1.65  =  2.99  volts,  practically  3. 


TABLKSo 


TABLES    OF  WEIGHTS. 

TABLES   OF   WEIGHTS.* 


176 


Milligram 

Gram 

Gram 

Gram 

Kilogram 

Kilogram 


METRIC  SYSTEM. 

0.0154323564  grain. 
15.4323564  grains. 
0.035273957  ounces  avoirdupois 
0.03215074  ounces  troy. 
2.20462234  pounds  avoirdupois. 
2.67922854  pounds  troy. 


AVOIRDUPOIS. 

Long  ton       ==       2240  pounds       =     1016.047  kilograms. 
Short  ton      =        2000  pounds       =      907.1849  kilograms. 
Pound    =    16  ounces    ==    7000  grains    =    453.5924277  grams. 
Ounce  =     437.5  grains  =      28.349527  grams. 

Grain  =       64.7989  milligrams    =        0.0647989  gram. 

TROY. 

Pound  =  12  ounces     =     5760  grains  =    373.24177  grams. 
Ounce  =  20  pennyweights  =  480  grains  =  31.103481  grams. 
Pennyweight  =  24  grains  =  1.555174  grams. 

Grain   =    64.7989  milligrams         =  0.0647989  gram. 

TROY  (Pharmacy). 
Ounce      =    8    drams      =    480  grains     =     31.103481  grams. 

Dram       =    3    scruples  ==      60  grains     =       3.887935  grams. 

Scruple   =  20  grains      =     1.295978  grams. 


*  United  States  Bureau  of  Standards,  September,  1904. 


176 


TABLES   OF   MEASURES. 


TABLES   OF   MEASURES.* 


Millimeter 

Centimeter 

Decimeter 

Meter 

Meter 

Meter 

Inch 

Foot  =12  inches. 

Yard  =  3  feet 

Mile  =  1760  yards 

Mile  =  1.609347  kilometers 


Square  millimeter 
''       centimeter 
"       decimeter 
"       decimeter 
"       meter 


NGTH. 

= 

0.03937  inch 

= 

0.3937  inch 

= 

3.937  inches 

= 

39.37  inches 

=: 

3.28083  feet 

= 

1.093611  yards 

= 

2.540005  centimeters 

= 

3.048006  decimeters 

=: 

0.914402  meter 

= 

5280  feet 

= 

1609.347  meters 

RFACE. 

0.00155    square    inch 

= 

0.1549997       '' 

= 

15.49997       ''       inches. 

= 

0.1076387      ''          foot. 

= 

1549.997       ''       inches. 

meter    =    10.76387  square  feet    =    1.19599 


yards. 


VOLUME. 

=  231  cubic  inches. 

=  3.7854345  liters. 

=  0.946359  liter. 

=  0.473179  liter. 

2.11336  pints  U.  S.        =         1.05668  quarts  U.  S. 

=  0.264170  gallon  U.  S. 
=      1.30794  cubic  yards      =      35.3145  cubic  feet. 
An  imperial  gallon,  English       =    4.5459631  liters 
=  277.410  cubic  inches  (U.  S.)=  277.412  cubic  inches  (Brit.). 


Gallon  U.  S 

Gallon     *' 

Quart      " 

Pint         " 

Liter        = 

Cubic  meter 

*  United  States  Bureau  of  Standards,  September,  1904. 


a 


INTERNATIONAL  ATOMIC   WEIGHTS. 

1904. 
INTERNATIONAL  ATOMIC  WEIGHTS. 


177 


0  =  16. 

H  =  l. 

0 

Aluminium  ... 

Al 

27.1 

26.9 

Neodymium  .... 

Nd 

Antimony 

Sb 

120.2 

119.3 

Neon 

....Ne 

Argon 

A 

As 

39.9 
75 

39.6 
74.4 

Nickel 

...  Ni 

Arsenic  

Nitrogen 

....N 

Barium 

Ba 

137.4 

136.4 

Osmium  

.....Os 

Bismuth 

Bi 

208.5 

206.9 

Oxygen 

....0 

Boron 

B 

11 

10.9 

Palladium 

....Pd 

Bromine 

....  Br 

79.96 

79.36 

Phosphorus 

....P 

Cadmium  .... 

....Cd 

112.4 

in.6 

Platinum 

Pt 

Caesium  

Cs 

132.9 

131.9 

Potassium 

....K 

Calcium 

Ca 

40.1 

39.8 

Praseodymium  . . 

....Pr 

Carbon    

C 

12 

11.91 

Radium 

....Rd 

Cerium 

Ce 

140.25 

139.2 

Rhodium  

....Rh 

Chlorine 

CI 

35.45 

35.18 

Rubidium 

...  Rb 

Chromium  .... 

Cr 

52.1 

5L7 

Ruthenium 

Ru 

Cobalt 

...Co 

59 

58.56 

Samarium 

Sm 

Columbium  ... 

Cb 

94 

93.3 

Scandium 

....Sc 

Copper  

....Cu 

63.6 

63.1 

Selenium 

Se 

Erbium  

....Er 

166 

164.8 

Silicon  

....Si 

Fluorine 

...  F 

19 

18.9 

Silver 

....Ag 

Gadolinium  ... 

Gd 

156 

155 

Sodium 

....Na 

Gallium 

....Ga 

70 

69.5 

Strontium 

...Sr 

Germanium  ... 

Ge 

72.5 

71.9 

Sulphur 

....S 

....Gl 
....Au 

9.1 
197.2 

9.03 
195.7 

Tantalum 

....Ta 

Gold 

Tellurium 

...Te 

....He 
H 

4 

1.008 
114 

4 

1 

113.1 

Terbium 

....Tb 

Hydrogen  

Indium 

Thallium 

....Tl 

....In 

Thorium 

....Th 

Iodine  

I 

126.85 

125.90 

Thulium 

....Tm 

Iridium  

....Ir 

193 

19L5 

Tin 

....Sn 

Iron 

....Fe 

55.9 

55.5 

Titanium  

...Ti 

Krypton 

....Kr 

8L8 

8L2 

Tungsten  

...W 

Lanthanum  ... 

....La 

138.9 

137.9 

Uranium 

...U       ! 

Lead  

....Pb 

206.9 

205.35 

Vanadium 

...V 

Lithium 

....Li 

7.03 

6.98 

Xenon 

...Xe     ] 

Magnesium.... 

....Mg 

24.36 

24.18 

Ytterbium 

...Yb     ] 

Manganese. ..„. 

....Mn 

55 

54.6 

Yttrium 

...Yt 

Mercury 

...Hg 

200 

198.5 

Zinc 

...Zn 

Molybdenum . . 

....Mo 

96 

95.3 

Zirconium 

...Zr 

)  =  16. 

H  =  l. 

143.6 

142.5 

20 

19.9 

58.7 

58.3 

14.04 

13.93 

191 

189.6 

16 

15.88 

106.5 

105.7 

31 

30.77 

194.8 

193.3 

39.15 

38.86 

140.5 

139.4 

225 

223.3 

103 

102.2 

85.4 

84.8 

10L7 

100.9 

150 

148.9 

44.1 

43.8 

79.2 

78.6 

28.4 

28.2 

107.93 

107.12 

23.05 

22.88 

87.6 

86.94 

32.06 

31.83 

183 

181.6 

127.6 

126.6 

160 

158.8 

204.1 

202.6 

232.5 

230.8 

171 

169.7 

119 

118.1 

48.1 

47.7 

184 

182.6 

238.5 

236.7 

5L2 

50.8 

128 

127 

173 

17L7 

89 

88.3 

65.4 

64.9 

90.6 

89.9 

178 


TABLE   OF   FACTORS. 

TABLE  OF   FACTORS. 


Required. 

Factor. 

Logarithm. 

A1203. 

Al. 

0.53033 

r.72454 

-  A1P04. 

Al. 

0.22195 

1.34626 

AI2O3. 

0.41851 

1.62171 

Sb204. 

Sb. 

0.78975 

1.89749 

Sb2S3. 

Sb. 

0.71424 

1.85384 

AS2S3. 

As. 

0.60931 

1.78484 

Mg2AS207. 

As. 

0.48275 

0.68373 

AgsAsOi. 

As. 

0.16206 

1.20968 

BaS04. 

BaO. 

0.65709 

1.81762 

SO3. 

0.34293 

1.53521 

S. 

0.13733 

1.13775 

Bi203. 

Bi. 

0.89678 

1.95269 

CdS. 

Cd. 

0.77807 

1.89102 

CaCOa. 

CO2. 

0.43956 

1.64302 

CaO. 

0.56044 

1.74853 

CaF2. 

F. 

0.48656 

1.68713 

CaSOi. 

CaO. 

0.41201 

1.61491 

CaC03. 

0.73515 

1.86638 

CO2. 

C. 

0.27273 

1.43573 

Cr203. 

Cr. 

0.68464 

1.83546 

8K2S04,2CoS04. 

Co. 

0.14162 

1.15113 

-      CuO. 

Cu. 

0.79900 

1.90255 

CU2S. 

Cu. 

0.79870 

1.90238 

Fe203. 

Fe. 

0.69962 

1.84486 

Fe. 

Fe203. 

1.42937 

0.15514 

FeO. 

1.28623 

0.10932 

Fe. 

Fe304. 

1.38167 

0.14040 

PbCrO^. 

Pb. 

0.64056 

1.80656 

PbS04. 

Pb. 

0.68294 

1.83438 

Li3l»04. 

Li20. 

0.38841 

1.58929 

LiCl. 

1.09777 

0.04051 

Mg2P207. 

P. 

0.278375 

1.44463 

P2O5. 

0.63757 

1.80453 

MgO. 

0.362425 

r.  55922 

TABLE   OF   FACTORS. 

TABLE  OF  FACTORS. 


179 


Required. 

Factor. 

Logarithm. 

Mg2P20, 

MgCOa. 

0.75755 

r.87941 

Mn304. 

Mn. 

0.72052 

1.85764 

Mn2P207. 

Mn. 

0.38732 

1.58807 

MnS. 

Mn. 

0.63174 

1.80054 

MnS04. 

Mn. 

0.36409 

1.56121 

HgS. 

Hg. 

0.86183 

1.93543 

NiO. 

Ni. 

0.78582 

1.89532 

(NH4)2PtCl6. 

Pt. 

0.43910 

1.64256 

N 

0.063295 

2.80137 

NH3. 

0.076927 

2.88608 

NH4CI. 

0.24128 

1.38253 

Pt  from(NH4)2PtCl6. 

N. 

0.14415 

1.15881 

NH3. 

0.17519 

1.24352 

NH4CI. 

0.54951 

1.73998 

KaPtCle. 

KCl. 

0.30712 

1.48731 

K2O. 

0.19411 

1.28805 

KCl. 

K2O. 

0.63203 

1.80074 

K2SO4. 

K2O. 

0.54083 

1.73306 

K. 

0.44907 

1.65231 

SiOa. 

Si. 

0.47020 

1.67228 

AgBr. 

Br. 

0.42556 

1.62896 

Agl. 

I. 

0.54030 

1.73263 

AgCl. 

CI. 

0.247245 

1.39313 

Ag. 

0.75275 

1.87665 

NaCl. 

Na20. 

0.53076 

1.72490 

Na2S04. 

Na20. 

0.43683 

1.64031 

Na. 

0.32428 

1.51092 

SrS04. 

SrO. 

0.56409 

1.75135 

Sn02. 

Sn. 

0.78808 

1.89657 

Ti02. 

Ti. 

0.60051 

1.77852 

WO3. 

W. 

0.79310 

1.89933 

ZnO. 

Zn. 

0.80345 

1.90496 

Zn2P207. 

Zn. 

0.42914 

1.63260 

ZnNH4P04. 

Zn. 

0.36644 

1.56400 

In  most  cases  the  factor  to  four  places  is  sufficient. 


180 


SPECIFIC    HEATS. 

SPECIFIC  HEATS. 


Lithium, 0.941 

Carbon, 0.463 

Sodium, 0.273 

Magnesium, 0.245 

Aluminum, 0.225 

Silicon, 0.203 

Phosphorus, 0.202 

Sulphur, 0.178 

Potassium, 0.166 

Calcium, 0.169 

Titanium, 0.148 

Manganese, 0.122 

Iron, 0.112 

Nickel, 0.108 

Cobalt, 0.107 

Chromium, 0.121 


Copper, 0.0930 

Zinc, 0.0935 

Arsenic, 0.0830 

Molybdenum, 0.0659 

Silver, 0.0570 

Cadmium, 0.0567 

Tin, 0.0540 

Antimony, 0.0523 

Tungsten,  ..,,,..  0.0350 
Mercury,  ..,,,,.  0.0328 

Platinum,  .  , 0.0324 

Gold, 0.0324 

Iridium, 0.0323 

Lead, 0.0307 

Bismuth, 0.0305 

Uranium, 0.0276 


Thermometers. 

Three  scales  are  now  in  general  use.     These  are  : 

1.  Centigrade— C.     Water  freezes  at    0°,  boils  at  100°. 

2.  Fahrenheit— F.         ''  ''  32°,       ''        212°. 


3.   Reaumur — R. 


To  Convert— F.  to  C. 


5(F.°— 32°) 


0^ 


=c.< 


80°. 


op  ° 

C.  toF.   ^+32°=F.° 
5 


RtoF.   ^+32= 


f: 


>■  Formulae. 


COMPARISON  OF  CENTIGRADE  AND  FAHRENHEIT  DEGREES.  181 

COMPARISON  OF  CENTIGRADE  AND  FAHRENHEIT  DEGREES. 


c° 

F° 

C° 

F° 

C° 

48 

F° 

C° 

F° 

C° 

F° 

—40 

—40 

22 

71.6 

118.4 

74 

165.2 

100 

212 

—30 

—22 

23 

73.4 

49 

120.2 

75 

167 

150 

302 

—20 

—  4 

24 

75.2 

50 

122 

76 

168.8 

200 

392 

—10 

14 

25 

77 

51 

123.8 

77 

170.6 

250 

482 

0 

32 

26 

78.8 

52 

125.6 

78 

172.4 

300 

672 

1 

33.8 

27 

80.6 

53 

127.4 

79 

174.2 

400 

752 

2 

35.6 

28 

82.4 

54 

129.2 

80 

176 

500 

932 

3 

37.4 

29 

84.2 

55 

131 

81 

177.8 

600 

1112 

4 

39.2 

30 

86 

56 

132.8 

82 

179.6 

700 

1292 

5 

41 

31 

87.8 

57 

134.6 

83 

181.4 

800 

1472 

6 

42.8 

32 

89.6 

58 

136.4 

84 

183.2 

900 

1652 

7 

44.6 

33 

91.4 

59 

138.2 

85 

185 

1000 

1832 

8 

46.4 

34 

93.2 

60 

140 

86 

186.8 

1100 

2012 

9 

48.2 

35 

95 

61 

141.8 

87 

188.6 

1200 

2192 

10 

50 

36 

96.8 

62 

143.6 

88 

190.4 

1300 

237^ 

11 

51.8 

37 

98.6 

63 

145.4 

89 

192.2 

1400 

2552. 

12 

53.6 

38 

100.4 

64 

147.2 

90 

194 

1500 

273^ 

13 

55.4 

39 

102.2 

65 

149 

91 

195.8 

1600 

2912 

14 

57.2 

40 

104 

6(i 

1 

150.8 

92 

197.6 

1700 

3092 

15 

59 

41 

105.8 

67 

152.6 

93 

199.4 

1800 

3272 

16 

60.8 

42 

107.6 

68 

154.4 

94 

201.2 

1900 

3452 

17 

62.6 

43 

109.4 

69 

156.2 

95 

203 

2000 

3632 

18 

64.4 

44 

111.2 

70 

158 

96 

204.8 

2500 

4532 

19 

66.2 

45 

113 

71 

159.8 

97 

206.6 

3000 

54355 

20 

68 

46 

114.8 

72 

161.6 

98 

208.4 

21 

69.8 

47 

116.6 

73 

163.4 

99 

210.2 

182 


TABLE   OF   VALUES   OF   NORMAL   SOLUTIONS. 


TABLE  OF  VALUES  OF   NORMAL   SOLUTIONS. 


Name.    • 

Formula. 

Molecular 
Weight. 

Weight  in 

Grams 
Per  Liter. 

Value  in 

Grams   of 

1.  C.  C. 

Sodium  Oxide 

Xa20 

62.1 

31.05 

0.03105 

Sodium  Hydroxide 

NaOH 

40.06 

40.06 

0.04006 

Sodium  Carbonate 

NasCOa 

106.1 

53.05 

0.05305 

Sodium  Bicarbonate 

NaHCOs 

84.06 

84.06 

0.08406 

Potassium  Oxide 

K2O 

94.3 

47.15 

0.04715 

Potassium  Hydroxide 

KOH 

56.16 

66.16 

0.05616 

Potassium  Carbonate 

K2CO3 

138.3 

69.15 

0.06915 

Potassium  Bicarbonate 

KHCO3 

100.16 

100.16 

0.10016 

Ammonia 

NH3 

17.06 

17.06 

0.01706 

Ammonium  Chloride 

NH4CI 

53.52 

53.52 

0.05352 

Calcium  Oxide 

CaO 

56.1 

28.05 

0.02805 

Calcium  Carbonate 

CaCOs 

100.1 

60.05 

0.05005 

Calcium  Sulphate 

CaS04 

136.16 

68.08 

0.06808 

Barium  Hydroxide 

Ba(0H)2 

171.42 

85.71 

0.08571 

Barium  Carbonate 

BaCOs 

197.4 

98.7 

0.0987 

Barium  Chloride 

BaCl2 

208.3 

104.15 

0.10415 

Magnesium  Oxide 

MgO 

40.36 

20.18 

0.02018 

Sodium  Sulphate 

Na2S04 

142.16 

71.08 

0.07108 

Potassium  Sulphate 

K2SO4 

174.36 

87.18 

0.08718 

Sulphuric  Acid 

H2SO4 

98.076 

49.038 

0.049038 

Nitric  Acid 

HNO3 

63.05 

63.05 

0.06305 

Hydrochloric  Acid. 

HCl 

36.46 

36.46 

0.03646 

Oxalic  Acid 

H2C2O4 

90.02 

45.01 

0.04501 

This  and  the  following  table  are  based  on  Oxygen   16.  and 
the  atomic  weights  given  on  page  477. 


VALUES   OF   TENTH    NORMAL   SOLUTIONS. 


183 


VALUES  OF  TENTH  NORMAL  SOLUTIONS. 


Name 


Potassium  Permanganate 

Iron 

Ferric  Oxide 

Ferrous  Oxide 

Magnetic  Oxide 

Ferrous  Sulphate  (Cryst) 

Ferrous  Ammonium  Sulphate 

Oxalic  Acid  (Cryst) 

Ammonium  Oxalate  (Cryst). . 

Calcium  Oxide 

Calcium  Carbonate 

Calcium  Sulphate  (Cryst) 

Manganese  (Volhard) 

Manganese  (Ford-Williams), . 

Phosphorus  (Noyes) 

Phosphoric  Oxide  (Noyes). . . 

Tin  from  SnClj 

Hydrogen  Sulphide 

Hydrogen  Peroxide 

Iodine 

Copper 

Antimony 

Sulphurous  Acid . , 

Sulphur  Dioxide 

Sodium  Thiosulphate 

Sodium  Thiosulphate  (Cryst). 

Chromium 

Chromic  Oxide 

Potassium  Bichromate 

Chlorine 

Potassium  Chloride 

Sodium  Chloride 

Ammonium  Chloride 

Silver 

Silver  Nitrate 


0^^ 

«  a  •? 

Formula. 

sS-? 

KMnO, 

158.15 

Fe. 

55.9 

FejO, 

159.8 

FeO 

7L9 

FeaO^ 

231.7 

FeSO^.  7H2O 

278.07 

FeSO^.  (NHJ2SO4.  6H2O 

392.26 

H2C2O4.  2H2O 

126.05 

(NHJ^CP,.  HP 

142.16 

CaO 

56.1 

CaCOg 

100.1 

CaS04.  SHjO 

172.19 

Mn. 

55 

Mn. 

55 

P. 

31 

P,05 

142 

Sn 

119 

HaS 

34.08 

H5O3 

34.02 

I 

126.85 

Cu 

63.6 

Sb 

120.2 

HjSO, 

82.08 

SOj 

64.06 

Na,S203 

158.22 

NajSjOg,  SHjO 

248.3 

Cr 

52.1 

Cr,03 

152.2 

K,Cr,0, 

294.5 

CI 

35.45 

KCl 

74.6 

NaCl 

58.5 

NH4CI 

Ag 
AgNO, 

53.52 
107.93 

169.97 

Weight  in 

Grams 
per  Liter. 


3.163 
5.59 
7.99 
7.19 
7.723 
27.807 
39.226 
6.3025 
7.108 
2.805 
5.005 
8.6095 
1.65 
2.75 
0.08609 
0.1972 
5.95 
1.704 
1.701 
12.685 
6.36 
6.01 
4.104 
3.203 
15.822 
24.83 
1.738 
2.537 
4.9083 
3.545 
7.46 
5.85 
5.352 
10.793 
16.997 


Value  in 
Grms.  of 
One  C.C. 


0.003163 

0.005590 

0.007990 

0.007190 

0.007723 

0.027807 

0.039226 

0.0063025 

0.007108 

0.002805 

0.005005 

0.0086095 

0.00165 

0.00275 

0.00008609 

0.0001972 

0.005950 

0.001704 

0.001701 

0.012685 

0.00636 

0.006010 

0.004104 

0.003203 

0.015822 

0.024830 

0.001738 

0.002537 

0.004908 

0.003545 

0.007460 

0.00585 

0.005352 

0.010793 

0.016997 


184 


DEGREES   BAUME   AND   SPECIFIC   GRAVITY. 


Table  of  Comparison — Degrees  of  the  Baume  Hydrometer  and 
Specific  Gravity.  * 

TABLE   FOR  LIQUIDS  HEAVIER  THAN  WATER. 


Degrees 

Specific 

Degrees 

Specific 

Degrees 

Specific 

Baum^. ' 

Gravity. 

Baum^. 

Gravity. 

Baum^. 

Gravity. 

0 

1.000 

26 

1.206 

52 

1.520 

1 

1.007 

27 

1.216 

53 

1.535 

2 

1.013 

28 

1.226 

54 

1.551 

3 

1.020 

29 

1.236 

55 

1.567 

4 

1.027 

30 

1.246 

56 

1.583 

5 

1.034 

31 

1.256 

57 

1.600 

6 

1.041 

32 

1.267 

58 

1.617 

7 

1.048 

33 

1.277 

59 

1.634 

8 

1.056 

34 

1.288 

60 

1.652 

9 

J.  063 

35 

1.299 

61 

1.670 

10 

1.070 

36 

1.310 

62 

1.689 

11 

1.078 

37 

1.322 

63 

1.708 

13 

1.086 

38 

1.333 

64 

1.7^7 

13 

1.094 

39 

1.345 

65 

1.747 

14 

1.101 

40 

1.357 

66 

1.767 

15 

1.109 

41 

1.369 

67 

1.788 

16 

1.118 

42 

1.382 

68 

1.809 

17 

1.126 

43 

1.395 

69 

1.831 

.18 

1.134 

44 

1.407 

70 

1.854 

19 

1.143 

45 

1.421 

71 

1.877 

20 

1.152 

46 

1.434 

72 

1.900 

21 

1.160 

47 

1.448 

73 

1.924 

22 

1.169 

48 

1.462 

74 

1.949 

23 

1.178 

49 

1.476 

75 

1.974 

24 

1.188 

50 

1.490 

76 

2.000 

25 

1.197 

51 

1.505 

TABLE  FOR 

.  LIQUIDS  LIGHTER  THAN  WATER. 

Degrees 

Specific 

Degrees 

Specific 

Degrees 

Specific 

Baume. 

Gravity. 

Banm^. 

Gravity. 

Baume. 

Gravity. 

10 

1.000 

27 

.896 

44 

.811 

11 

.993 

28 

.890 

45 

.807 

12 

.986 

29 

.885 

46 

.802 

13 

.980 

30 

.880 

47 

.798 

14 

.973 

31 

.874 

48 

.794 

15 

.967 

32 

.869 

49 

.789 

16 

.960 

33 

.864 

50 

.785 

17 

.954 

34 

.859 

51 

.781 

18 

.948 

35 

.854 

52 

.777 

19 

.942 

36 

.849 

53 

.773 

20 

.936 

37 

.844 

54 

.768 

21 

.930 

3S 

.839 

55 

.764 

22 

.924 

39 

.834 

56 

.760 

23 

.918 

40 

.830 

57 

.757 

24 

.913 

41 

.825 

58 

.753 

25 

.907 

42 

.820 

59 

.749 

26 

.901 

43 

.816 

60 

.745 

*  Crooke's  Select  Methods,  p.  693. 


VOLUME  AND  WEIGHT  OF  WATER.  185 

VOLUME  AND  WEIGHT  OF  WATER  FROM  0°C  TO  31°C.^ 


Tempera- 

Volume of  one 

Weight  of  one 

Tempera- 

Volume of  one 

Weight  of  one 

ture. 

gram  in  c.c. 

c.c.  in  grams. 

ture. 

gram  in  c.c. 

c  c.  in  grams. 

0° 

1.000126 

0.999874 

16° 

1.001025 

0.998976 

1 

1.000070 

0.999930 

17 

1.001193 

0.998808 

2 

1.000030 

0.999970 

18 

1.001373 

0.998629 

3 

1.000007 

0.999993 

19 

1.001564 

0.998438 

4 

1.000000 

1.000000 

20 

1.001768 

0.998235 

5 

1.000008 

0.999992 

21 

1.001981 

0.998023 

6 

1.000031 

0.999969 

22 

1.002204 

0.997801 

7 

1.000069 

0.999931 

23 

1.002438 

0.997568 

8 

1.000122 

0.999878 

24 

1.002681 

0.997326 

9 

1.000188 

0.999812 

25 

1.002935 

0.997073 

10 

1.000269 

0.999731 

26 

1.003199 

0.996811 

11 

1.000363 

0.999637 

27 

1.003472 

0.996540 

12 

1.000470 

0.999530 

28 

1.003788 

0.996226 

13 

1.000590 

0.999410 

29 

1.004045 

0.995971 

14 

1.000722 

0.999278 

30 

1.004346 

0.995673 

15 

1.000867 

0.999134 

31 

1.004656 

0.995365 

For  the  most  recent  results  on  the  expansion  of  water,  which 
do  not  differ  materially  from  the  preceding,  see  Annalen  d. 
Physik  n.  Chemie  [N.  F.]  60,  340,  1897. 


Wied.  Ann.  47.400.  1892. 


186 


WEIGHTS  OF  GASES. 


Weights  of  One  Liter  of  Various  Gases  ato°  C.  and  760  m.m. 

Pressure  at  the  Sea  Level  in  45°  Latitude,  Calculated  from 

Morley's  Results.*    Hydrogen  0.08973  Gram. 

Oxygen  1.429  Grams. 


Name. 


Acetylene 

Allylene 

Ammonia 

Arsine 

Bromine  

Carbon  Dioxide 

Carbon  Monoxide.... 
Carbon  Oxysulphide 

Chlorine  

Cyanogen 

Ethane 

Ethylene 

Hydriodic  Acid 

Hydrobromic  Acid... 
Hydrochloric  Acid... 
Hydrofluoric  Acid  ... 

Hydrogen 

Hydrogen  Sulphide.. 

Methane 

Nitric  Oxide  

Nitrogen 

Nitrous  Oxide 

Oxygen  

Phosphine  

Silicon  Tetrafluoride. 

Sulphur  Dioxide 

Water  Vapor..  

Air 


Formula. 


NH3 

AsH, 

Br, 

CO, 

CO 

COS 

CI, 

(CN). 

C,H. 

HI 
HBr 
HCI 

HF 

H, 
H,S 
CH. 
NO 

N, 
N,0 

O, 
PH, 
SiF. 
SO, 
H,0 


Weight  of 
One  Liter. 


1.16178 
1.78768 
0.76201 
3.48426 
7.14139 
1.96487 
1.25037 
2.68205 
3.16613 
2.32570 
1.34183 
1.25180 
5.70966 
3.61573 
1.62808 
0.89348 
0.08973 
1.52171 
0.71593 
1.34147 
1.25395 
1.96845 
1.42900 
1.51938 
4.66211 
2.86068 
0.80453 
1.29327t 


These  numbers  are  also  the  weight  of  a  cubic  meter  in  Kilograms. 

*  Smithsonian  Contribution  980,  p.  no,  1895. 

f  This  value  is  taken  from  Travers'  Gases  of  the  Atmosphere. 


VAPOR   PRESSURE  OF  WATER. 


187 


VAPOR  PRESSURE  OF  WATER  FROM  0°  C  TO  100°  C 
IN  MILLIMETERS  OF  MERCURY.* 


Temp. 

Pressure. 

Temp. 

Pressure. 

Temp. 

Pressure. 

Temp. 

Pressure. 

0 

4.569 

25 

23.517 

50 

91.98 

75 

288.76 

1 

4.909 

26 

24.956 

51 

96.66 

76 

301.09 

2 

5.272 

27 

26.471 

52 

101.55 

77 

313.85 

3 

5.658 

28 

28.065 

53 

106.65 

78 

327.05 

4 

6.069 

29 

29.744 

54 

111.97 

79 

340.73 

5 

6.507 

30 

31.51 

55 

117.52 

80 

354.87 

6 

6.972 

31 

33.37 

56 

123.29 

81 

369.51 

7 

7.466 

32 

35.32 

57 

129.31 

82 

384.64 

8 

7.991 

33 

37.37 

58 

135.58 

83 

400.29 

9 

8.548 

34 

39.52 

59 

142.10 

84 

416.47 

10 

9.140 

35 

41.78 

60 

148.88 

85 

433.19 

11 

9.767 

36 

44.16 

61 

155.95 

86 

450.47 

12 

10.432 

37 

46.65 

62 

163.29 

87 

468.32 

13 

11.137 

38 

49.26 

63 

170.02 

88 

486.76 

14 

11.884 

39 

52.00 

64 

178.86 

89 

505.81 

15 

12.674 

40 

54  87 

65 

187.10 

90 

525.47 

16 

13.510 

41 

57  87 

66 

195.67 

91 

545.77 

17 

14.395 

42 

6L02 

67 

204.56 

92 

566.71 

18 

15.330 

43 

64.31 

68 

213.79 

93 

588.83 

19 

16.319 

44 

67.76 

69 

223.37 

94 

610.64 

20 

17.363 

45 

71.36 

70 

233.31 

95 

633.66 

21 

18.466 

46 

75.13 

71 

243.62 

96 

657.40 

22 

19.630 

47 

79.07 

72 

254.30 

97 

681.88 

23 

20.858 

48 

83.19 

73 

265.38 

98 

707.13 

24 

22.152 

49 

87  49 

74 

276.87 

99 

733.16 

*  Taken  from  Ostwald's  Manual  of  Physico-Chemical  Measurements. 


188 


SPECIFIC    GRAVITY    OF  ALCOHOL. 


TABLE   OF  SPECIFIC   GRAVITY  AND   PERCENTAGE  OF 
ALCOHOL   (ETHYL).* 


Per  Cent. 

Per  Cent. 

Specific 

Per  Cent. 

Per  Cent 

Specific 

by 

by 

Gravity  at 

i5.56°C. 

by 

by 

Gravity  at 

Volume. 

Weight. 

Volume. 

Weight. 

i5.56°C. 

0 

0.00 

1.0000 

51 

43.47 

.9315 

1 

0.80 

.9976 

52 

44.42 

.9295 

2 

1.60 

.9961 

53 

45.86 

.9275 

3 

2.40 

.9947 

54 

46.32 

.9254 

4 

3.20 

.9933 

55 

47.29 

.9234 

5 

4.00 

.9919 

56 

48.26 

.9213 

6 

4.81 

.9906 

57 

49.23 

.9192 

7 

5.62 

.9893 

58 

50.21 

.9170 

8 

6.43 

.9881 

59 

51.20 

.9148 

9 

7.24 

.9869 

60 

52.20 

.9126 

10 

8.05 

.9857 

61 

53.20 

.9104 

11 

8.87 

.9845 

62 

54.21 

.9082 

12 

9.69 

.9834 

63 

55.21 

.9059 

13 

10.51 

.9823 

64 

56.22 

.9036 

14 

11.33 

.9812 

65 

57.24 

.9013 

15 

12.15 

.9802 

66 

58.27 

.8989 

16 

12.98 

.9791 

67 

59.82 

.8965 

17 

13.80 

.9781 

68 

60.38 

.8941 

18 

14.63 

.9771 

69 

61.42 

.8917 

19 

15.46 

.9761 

70 

62.50 

.8892 

20 

16.28 

.9751 

71 

63.58 

.8867 

21 

17.11 

.9741 

72 

64.66 

.8842 

22 

17.95 

.9731 

73 

65.74 

.8817 

23 

18.78 

.9720 

74 

66.83 

.8791 

24 

19.62 

.9710 

75 

67.93 

.8765 

25 

20.46 

.9700 

76 

69.05 

.8739 

26 

21.30 

.9689 

77 

70.18 

.8712 

27 

22.14 

.9679 

78 

71.31 

.8685 

28 

22.99 

.9668 

79 

72.45 

.8658 

29 

23.84 

.9657 

80 

73.59 

.8631 

30 

24.69 

.9646 

81 

74.74 

.8603 

31 

25.55 

.9634 

82 

75.91 

.8575 

32 

26.41 

.9622 

83 

77.09 

.8547 

33 

27-.  27 

.9609 

84 

78.29 

.8518 

34 

28.13 

.9596 

85 

79.50 

.8488 

35 

28.99 

.9583 

86 

80.71 

.8458 

36 

29.86 

.9570 

87 

81.94 

.8428 

87 

30.74 

.9556 

88 

83.19 

.8397 

38 

31.62 

.9541 

89 

84.46 

.8365 

39 

32.50 

.9526 

90 

85.75 

.8332 

40 

33.39 

.9510 

91 

87.09 

.8299 

41 

34.28 

.9494 

92 

83.87 

.8265 

42 

35.18 

.9478 

93 

89.71 

.8230 

43 

36.08 

.9461 

94 

91.07 

.8194 

44 

36.99 

.9444 

95 

92.46 

.8157 

45 

37.90 

.9427 

93 

98.89 

.8118 

46 

38.82 

.9409 

97 

95.34 

.8077 

47 

39.75 

.9391 

98 

96.84 

.8034 

48 

40.66 

.9373 

99 

98.39 

.7988 

49 

41.59 

.9354 

100 

100.00 

.7989 

50 

42.52 

.9335 

*  Tralles. 


SPECIFIC  GRAVITY  OF  SULPHURIC  ACID. 


189 


PERCENTAGE    AND    SPECIFIC   GRAVITY  OF    SULPHURIC 
ACID  AT   15°  C*     WATER   AT    0°  C.  =  1. 


Percent. 

Sp.  Gr. 

Percent. 

Sp.  Gr. 

Percfent. 

Sp.  Gr. 

Percent. 

Sp.  Gr. 

1 

1.006 

26 

1.191 

51 

1.409 

76 

1.684 

2 

1.012 

27 

1.199 

52 

1.418 

77 

1.697 

3 

1.018 

28 

1.207 

53 

1.428 

78 

1.710 

4 

1.025 

29 

1.215 

54 

1.438 

79 

1.721 

5 

1.032 

30 

1.223 

55 

1.448 

80 

1.732 

6 

1.039 

31 

1.231 

56 

1.459 

81 

1.743 

7 

1.046 

32 

1.239 

57 

1.469 

82 

1.753 

8 

1.053 

33 

1.247 

58 

1.480 

83 

1.763 

9 

1.061 

34 

1.256 

59 

1.491 

84 

1.773 

10 

1.069 

35 

1.264 

60 

1.501 

85 

1.783 

11 

1.076 

36 

1.272 

61 

1.512 

86 

1.792 

12 

1.084 

37 

1.281 

62 

1.523 

87 

1.800 

13 

1.091 

38 

1.290 

63 

1.535 

88 

1.807 

14 

1.099 

39 

1.298 

64 

1.546 

89 

1.814 

15 

1.106 

40 

1.307 

65 

1.558 

90 

1.820 

16 

1.114 

41 

1.316 

66 

1.569 

91 

1.825 

17 

1.122 

42 

1.324 

67 

1.580 

92 

1.8294 

18 

1.129 

43 

1.333 

68 

1.592 

93 

1.8339 

19 

1.137 

44 

1.342 

69 

1.604 

94 

1.8372 

20 

1.145 

45 

1.352 

70 

1.615 

95 

1.8390 

21 

1.153 

46 

1.361 

71 

1.626 

96 

1.8406 

22 

1.161 

47 

1.370 

72 

1.638 

97 

1.8410 

23 

1.168 

48 

1.379 

73 

1.650 

98 

1.8412 

24 

1.176 

49 

1.389 

74 

1.662 

99 

1.8403 

25 

1.184 

50 

1.399 

75 

1.674 

100 

1.8384 

*  Gerlach  Z.  Anal.  Chem.  27,316. 


190 


SPECIFIC   GRAVITY   OF  NITRIC   ACID. 


PERCENTAGE  AND   SPECIFIC  GRAVITY   OF   NITRIC   ACID 

AT  15°C.  * 


Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

1 

1.006 

26 

1.159 

51 

1.323 

76 

1.445 

2 

1.012 

27 

1.166 

52 

1.329 

77 

1.449 

3 

1.018 

28 

1.172 

53 

1.335 

78 

1.452 

4 

1.024 

29 

1.179 

54 

1.341 

79 

1.456 

5 

1.029 

30 

1.185 

55 

1.346 

80 

1.460 

6 

1.035 

31 

1.192 

56 

1.352 

81 

1.463 

7 

1.010 

32 

1.198 

57 

1.358 

82 

1.467 

8 

1.045 

33 

1.204 

58 

1.363 

83 

1.470 

9 

1.051 

34 

1.210 

69 

1.369 

84 

1.474 

10 

1.057 

35 

1.218 

60 

1.374 

85 

1.478 

11 

1.064 

36 

1.225 

61 

1.380 

86 

1.481 

12 

1.070 

37 

1.230 

62 

1.386 

87 

1.484 

13 

1.077 

38 

1.236 

63 

1.390 

88 

1.488 

14 

1.083 

39 

1.244 

64 

1.395 

89 

1.491 

15 

1.089 

40 

1.251 

65 

1.400 

90 

1.495 

16 

1.095 

41 

1.257 

66 

1.405 

91 

1.499 

17 

1.100 

42 

1.264 

67 

1.410 

92 

1.503 

18 

1.106 

43 

1.270 

68 

1.414 

93 

1.506 

19 

1.112 

44 

1.276 

69 

1.419 

94 

1.509 

20 

1.120 

45 

1.284 

70 

1.423 

95 

1.512 

21 

1.126 

46 

1.290 

71 

1.427 

96 

1.516 

22 

1.132 

47 

1.298 

72 

1.431 

97 

1.520 

23 

1.138 

48 

1.304 

73 

1.435 

98 

1.523 

24 

1.145 

49 

1.312 

74 

1.439 

99 

1.526 

25 

1.151 

50 

1.316 

75 

1.442 

100 

1.530 

*  Kolb,  Gerlach,  Z.  Anal.  Chem.  8,292. 


SPECIFIC  GRAVITY  OF  HYDROCHLORIC  ACID. 


191 


SPECIFIC  GRAVITY  AND    PERCENTAGE   OF   HYDRO- 
CHLORIC ACID  AT  15°  C*    WATER  AT  4°  C  =  1. 


Sp.  Gr. 

%nc\. 

Sp.  Gr. 

%  HCl. 

1.000 

0.16 

1.105 

20.97 

1.005 

1.15 

1.110 

21.92 

1.010 

2.14 

1.115 

22.86 

1.015 

3.12 

1.120 

23.82 

1.020 

4.13 

1.125 

24.78 

1.025 

5.15 

1.130 

25.75 

1.030 

6.15 

1.135 

26.70 

1.035 

7.15 

1.140 

27.66 

1.040 

8.16 

1.145 

28.61 

1.045 

9.16 

1.150 

29.57 

1.050 

10.17 

1.155 

30.55 

1.055 

11.18 

1.160 

31.52 

1.060 

12.19 

1.165 

32.49 

1.065 

13.19 

1.170 

33.46 

1.070 

14.17 

1.175 

34.42 

1.075 

15.16 

1.180 

35.39 

1.080 

16.15 

1.185 

36.31 

1.085 

17.13 

1.190 

37.23 

1.090 

18.11 

1.195 

38.16 

1.095 

19.06 

1.200 

39.11 

1.100 

20.01 

For  other  similar  tables  see  Comey's  Dictionary  of  Chemical 

Solubilities. 

*  Lunge  and  Marchlewski  Zeit.  f.  angew.  Chem.  1891.  183. 


192 


SPECIFIC   GRAVITY   OF   AMMONIA. 


SPECIFIC   GRAVITY   AND   PERCENTAGE   OF 
AMMONIA*   AT   15°    C. 


Sp.  Gr. 

%NH3 

Sp.  Gr. 

%NH, 

Sp.  Gr. 

^NH, 

1.000 

0.00 

0.960 

9.91 

0.920 

21.75 

0.998 

0.45 

0.958 

10.47 

0.918 

22.39 

0.996 

0.91 

0.956 

11.03 

0.916 

23.03 

0.994 

1.37 

0.954 

11.60 

0.914 

23.68 

0.992 

1.84 

0.952 

12.17 

0.912 

24.33 

0.990 

2.31 

0.950 

12.74 

0.910 

24.99 

0.988 

2.80 

0.948 

13.31 

0.908 

25.65 

0.986 

3  30 

0.946 

13.88 

0.906 

26.31 

0.984 

3.80 

0.944 

14.46 

0.904 

26.98 

0.98a 

4.30 

0.942 

15.04 

0.902 

27.65 

0.980 

4.80 

0.940 

15.63 

0.900 

28.33 

0.978 

5.30 

0.938 

16.22 

0  898 

29.01 

0.976 

5.80 

0.936 

16.82 

0.896 

29.69 

0.974 

6.30 

0.934 

17.42 

0.894 

30.37 

0.972 

6.80 

0.932 

18.03 

0.892 

31.05 

0.970 

7.31 

0.930 

18.64 

0.890 

31.75 

0.968 

7.82 

0.928 

19.25 

0.888 

32.50 

0.966 

8.33 

0.926 

19.87 

0.886 

33.25 

0.964 

8.84 

0.924 

20.49 

0.884 

34.10 

0.962 

9.35 

0.922 

21.12 

0.882 

34.95 

Lunge  and  Wiernik  Zeit .  f ,  angew.  Chem.  1889.  183. 


ELECTRICAL  UNITS. 


193 


TABLE  SHOWING   THE   RELATIONS   BETWEEN   UNITS   OF 
ELECTRICITY,  HEAT  AND  POWER. 


1  ampere  ==  1  volt  -;-  1  ohm. 

=  1  coulomb  per  second. 
1  ampere  hour  ^  1  coulomb  per  second 
kept  up  for  one  hour. 
1  ampere  hour  ==3600  coulombs. 
1  volt  =  1  ampere  X  1  ohm. 
1  ohm  =  1  volt  -:-  1  ampere. 
1  joule  =  1  volt  X  1  coulomb. 
"      =  .2381  calorie. 
«'      =  .73732  foot-pound. 
"      =  .10194  kilogram-meter. 
1  calorie  =  4.2  joules. 

=  3.0968  foot-pounds. 
"       ==  .42815  kilogram-meter. 
1  foot  pound  =  1.3563  joules. 
=  .32292  calorie. 
"  ==  .13825  kilogram-meter 

1  kilogram  meter  =  9.81  joules. 

=  2.3362  calories. 
=  7.233  foot-pounds. 
1  watt-hour  =  1  watt  kept  up  for  one 

hour. 
1  watt-hour  =  1  joule  per  second  kept 

up  for  one  hour. 
1  watt-hour  =  3600  joules. 

==857.16  calories. 
••        =  2654.4  foot-pounds. 

=  366.98  kilogram-meters 
1  watt  =  1  joule  per  second. 

"     =  .2381  calorie  per  second. 
• '     =  .  73732  foot-pound  per  second. 


1  watt  ==  .10194  kilogram-meter   per 

second. 
1  watt  =  .0013406  horse-power. 

"     =  .001  kilowatt. 
1  horse-power  =  745.94  watts. 

"  =177.6  calories  per  sec- 

ond. 
1  horse-power  =  10656.    calories    per 

minute. 
1  horse -power  =  550  foot  pounds  per 

second. 
1  horse-power  =  33000  foot  pounds  per 

minute. 
1  horse-power  =  76.04  kilogram-met- 
ers per  second. 
1  horse-power  ==  4562. 4  kilogram-met- 
ers per  minute. 
1  horse-power  =  .74594  kilowatts. 
1  kilowatt  =  1000  watts. 

«'         =  1000  joules  per  second. 

"        =  238.1  calories  per  second. 

• '         =  14286.  calories  per  minute 

=  737.32  foot-pounds  per 

second. 

1  kilowatt  ==  44239.    foot-pounds  per 

minute. 
1  kilowatt  ■=  101.94    kilogram-meters 

per  second. 
1  kilowatt  =  6116.4    kilogram-meters 

per  minute. 
1  kilowatt  =  1.3406  horse-power. 


iU 


TABLE    OF    ELECTRO-CHEMICAL    EQUIVALENTS. 


Table  of  Electro-Chemical  Equivalents  based  on  the  definition  of  the 
ampere  and  the  atomic  weights,  oxygen  being  16,  as  given 
on  page  177. 


Element. 


Electropositive. 

Aluminium 

Antimony 

Bismuth 

Cadmium 

Cobalt 

Copper  (Cuprous) 

Copper  (Cupric) 

Gold 

Hydrogen 

Iron  (Ferrous) 

Iron  (Ferrie) 

Lead 

Magnesium 

Manganese 

Mercury  (Mercurous). . 
Mercury  (Mercuric) . . . 

Nickel 

Platinum 

Potassium. ....     .    . . 

Silver 

Sodium 

Tin  (Stannous) 

Tin  (Stannic) 

Zinc , 


Electronegative. 

Bromine 

Chlorine 

Iodine. ... 

Nitrogen 

Oxygea 


.5? 


27.1 

120.2 

208.5 

112.4 

59.0 

63.6 

63.6 

197.2 

1.008 

55.9 

55.9 

206.9 

24.36 

55.0 

200. 

200. 

58.7 

194.8 

39.15 

107.93 

23.05 

119.0 

119.0 

65.4 

79.96 
35.45 
126.85 
14.04 
16. 


c 

> 

c 
U 

Electro-chemical 

Equivalent. 
Grams  per 

coulomb. 

Si: 

is. 

o  s 

3. 

9.033 

.00009358 

10686. 

0.3369 

3. 

40.07 

.0004151 

2409. 

1.495 

3. 

69.5 

.0007200 

1389. 

2.592 

2. 

56.2 

.0005822 

1718. 

2.096 

2. 

29.5 

.0003056 

3272. 

1.100 

1. 

63.6 

.0006589 

1518. 

2.372 

2. 

3L8 

.0003295 

3036. 

1.186 

3. 

65.73 

.0006810 

1469. 

2.452 

1. 

1.008 

.00001044 

95785. 

.03758 

2. 

27.95 

.0002896 

3453. 

1.043 

3. 

18.63 

.0001930 

5181. 

0.6949 

2. 

103.45 

.0010717 

9331. 

3.859 

2. 

12.18 

.0001262 

7925. 

0.4543 

2. 

27.50 

.0002849 

3510. 

1.026 

1. 

200. 

.002072 

482.6 

7.459 

2. 

100. 

.001036 

965.3 

3.730 

2. 

29.35 

.0003041 

3288. 

1.095 

4. 

48.7 

.0005045 

1982. 

1.817 

1. 

39.15 

.0004056 

2466. 

1.460 

1. 

107.93 

.001118 

894.3 

4.026 

1. 

23.05 

.0002388 

4188. 

0.8597 

2. 

69.5 

.0006164 

1622. 

2.219 

4. 

29.75 

.0003082 

3245. 

LllO 

2. 

32.70 

.0003388 

2952. 

1.220 

1. 

79.96 

.0008284 

1207. 

2.982 

1. 

35.45 

.0003673 

2723. 

1.322 

1. 

126.85 

.001314 

761.0 

4.730 

3. 

4.68 

.00004848 

20627. 

0.1745 

2. 

8. 

.00008288 

12066. 

0.2984 

•  5 
i>  o 


1346. 

303.4 

174.9 

216.3 

412.2 

19L2 

382.4 

185.0 

12070. 

435.0 

652.6 

117.5 

998.1 

442.1 

60.81 

12L6 

414.2 

249.6 

310.5 

112.6 

527.6 

204.3 

408.6 

371.8 

152.0 
343.1 
95.90 
2599. 
1520. 


HEATS   OF   COMBINATION. 


195 


Heats  of   Combination    in   Calories,    for  Equivalent    Weights  in 
grams,  of  Chlorides,  Bromides,  Iodides,  Sulphates  and  Nitrates.* 


Element. 

Valence. 

Chloride. 

Bromide. 

Iodide. 

Sulphate. 

Nitrate. 

Aluminium 

3 

53660 

39900 

23463 

25315 

Antimony 

3 

30463 

Bismuth 

3 

30210 

Cadmium 

2 

46620 

37600 

24215 

44940 

43000 

Cobalt 

2 

38240 

44350 

42270 

Copper 

1 

32875 

24985 

16260 

Copper 

2 

27980 

26205 

Gold 

3 

7607 

2950 

Hydrogen 

1 

39315 

28380 

13170 

39170 

33830 

Iron 

2 

41025 

46600 

44835 

Lead 

2 

41385 

32225 

19900 

34035 

Magnesium 

2 

75505 

90090 

88240 

Manganese 

.2 

55995 

60625 

58860 

Mercury 

2 

31580 

17155 

18535 

Nickel 

2 

37265 

43475 

41710 

Silver 

1 

29380 

22700 

13800 

10195 

8390 

Tin 

2 

40395 

Zinc 

2 

48605 

37965 

24615 

53045 

51255 

*  These  values  are  taken  from  Thomsen's  Thermo-chemische 
Untersuchungen.  The  sulphates,  nitrates  and  hydrogen  com- 
pounds  are  for  aqueous  solutions.  The  chlorides,  bromides  and 
iodides  are  anhydrous. 


19G  SPECIFIC   RESISTANCE. 

SPECIFIC  RESISTANCE   OF  VARIOUS   SUBSTANCES 


Substance. 


Silver  (annealed) 

Silver  (hard  drawn) 

Copper  (annealed) 

Copper  (hard  drawn) 

Gold  (annealed) 

Gold  (hard  drawn) 

Aluminium  (annealed) 

Platinum  (annealed) 

Iron   (annealed) 

Tin  (pressed) 

Zinc  (pressed) 

Lead  (pressed) 

Nickel  (pressed) 

German  Silver 

Graphite 

Retort  Carbon 

Mercury , 

Nitric  Acid  in  water 

Hydrochloric  Acid  in  water. . . . 

Sulphuric  Acid  in  water 

Phosphoric  Acid  in  water 

Tartaric  Acid  in  water 

Acetic  Acid  in  water 

Ammonium  Chloride  in  water. . 

Sodium  Chloride  in  water 

Sodium  Sulphate  in  water 

Zinc  Sulphate  in  water 

Zinc  Sulphate  in  water 

Copper  Sulphate  in  water 

Potassium  Sulphate  in  water.  . . 
Potassium  Bichromate  in  water. 


4> 

5^ 

^ 

o   ^ 

^ 

S'^ 

4J 

H 

c^ 

0°C. 

10.6 

•* 

10.6 

*' 

8.9 

t( 

8.9 

(< 

19.3 

«< 

19.3 

" 

2.6 

" 

21.2 

'♦ 

8.1 

" 

7.3 

<i 

7.1 

«« 

11.4 

<< 

8.5 

14 

2.3 

•' 

1.9 

" 

13.6 

18°C. 

1.185 

«« 

1.092 

'« 

1.224 

«« 

1.307 

<< 

1.107 

" 

1.022 

10°C. 

<< 

1.270 

t( 

1.422 

<( 

<( 

1.205 

p 
in  ohms. 


.000001500 
.000001530 
.000001594 
.000001629 
.000002052 
.000002089 
.000002903 
.000009030 
.000009687 
.00001317 
.000005598 
.00001957 
.00001242 
about  .0000209 
0024  to  .042 

.07 
.000094073* 
1.28 
1.31 
1.36 
4.79 
9.97 
61.9 
2.5 
4.7 
11.3 
28.5 
33.7 
29.3 
16.6 
29.6 


29.7 
18.3 
30.4 
46.8 
22.4 
16.6 


Observer. 


Matthiesen. 


Everett. 


Kohlrausch. 


Kohlrausch 

and 
Nippoldt. 
Ewing  &  Macgregor 


Further  data  on  the  conductivity  of  solutions  can  be  found  in  Physikalisch-Chem- 
ische  Tabellen,  Landolt  and  Bornstein,  pp.  103,  106. 

Kohlrausch  and  Nippoldt,  Pogg.  Ann.  138  p.  379,  1869. 

Grotrian  "  "      151  p.  378,  1874. 

Kohlrausch  and  Grotrian,  Pogg.  Ann.  159  p.  233,  1876  and  Wied.  Ann.  6,  p.  145, 
1879. 

*  From  the  definition  of  the  ohm. 
f  Solution  of  minimum  resistance. 
X  Saturated  solution. 


COMPARISON  OF  VOLTAGES.  19 7 

Comparison  of  Calculated  and  Observed  Decomposition  Voltages  of 
various  Aqueous  Solutions. 


Chloride. 

Bromide. 

Iodide. 

Sulphate. 

Nitrate. 

1 

i 

i 

1 
1 

1 

1 

1 
0 

1 

1 
1 

1 

6 

1 

Magnesium. 

3.28 

3.1 

2.56 

2.01 

3.91 

3.83 

Zinc.    .  .   . 

2.11 

2.11 

1.65 

1.79 

1.07 

1.25 

2.31 

2.35 

2.23 

Cadmium.  . 

2.03 

1.9 

1.63 

1.58 

1.06 

1.12 

1.95 

2.03 

1.87 

1.98 

Aluminium. 

2.33 

2.0 

1.74 

1.53 

1.02 

.88 

1.10 

Iron.    .   .   . 

1.78 

1.6 

1.30 

.68 

2.03 

1.95 

Cobalt.    .   . 

1.66 

1.43 

1.05 

.51 

1.91 

1.92 

1.84 

Nickel.  .   .  . 

1.62 

1.33 

.85 

.36 

1.89 

2.09 

1.81 

Tin 

1.76 

1.61 

1.30 

.71 

1 

Lead.  .  .  . 

1.80 

1.63 

1.40 

1.33 

.87 

.83 

1.48 

1.52 

Copper.   .    . 

1.43 

1.32 

1.09 

1.02 

.71 

.64 

1.22 

1.14 

Silver.    .    . 

1.28 

1.11 

.99 

.95 

.60 

.65 

.44 

.365 

.36* 

Antimony.  . 

1.32 

1.22 

.80 

.44 

Bismuth.   . 

1.31 

1.21 

.92 

.43 

Hydrogen.  . 

1.71 

1.31 

1.23 

.94 

.57 

.52 

1.70 

1.67 

1.81 

1.69 

The  observed  decomposition  values  of  the  acids  given 
under  hydrogen  together  with  many  of  the  sulphates  and 
nitrates  are  taken  from  LeBlanc's  Electro  Chemistry  pp.  247-8. 
The  other  observed  values  are  from  Crocker,  Trans.  Am.  Inst. 
E.E.  1885,  p.  281. 

The  calculated  values   are  obtained  by  substituting  the 

TT 

heats  of  combination  on  page  177  in  equation  (14)  E  =  23000" 


Experiment. 


198 


LOGARITHMS   OF   NUMBERS. 


IJ 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Proportional  Parts. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

4 

8 

12 

17 

21 

25 

29 

33 

37 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

4 

8 

11 

15 

19 

23 

26 

30 

34 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

3 

7 

10 

14 

17 

21 

24 

28 

31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

3 

6 

10 

13 

16 

19 

23 

26 

29 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

3 

6 

9 

12 

15 

18 

21 

24 

27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

3 

6 

8 

11 

14 

17 

20 

22 

25 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

3 

5 

8 

11 

13 

16 

18 

21 

24 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

2 

5 

7 

10 

12 

15 

17 

20 

22 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

2 

5 

7 

9 

12 

14 

16 

19 

21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

2 

4 

7 

9 

11 

13 

16 

18 

20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

2 

4 

6 

8 

11 

13 

15 

17 

19 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

2 

4 

6 

8 

10 

12 

14 

16 

18 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

2 

4 

6 

8 

10 

12 

14 

15 

17 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

2 

4 

6 

7 

9 

11 

13 

15 

17 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

2 

4 

5 

7 

9 

11 

12 

14 

16 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

2 

3 

5 

7 

9 

10 

12 

14 

15 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

2 

3 

5 

7 

8 

10 

11 

13 

15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

2 

3 

5 

6 

8 

9 

11 

13 

14 

28 

4172 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

2 

3 

5 

6 

8 

9 

11 

12 

14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

1 

3 

4 

6 

7 

9 

10 

12 

13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

3 

4 

6 

7 

9 

10 

11 

13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

3 

4 

6 

7 

8 

10 

11 

12 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

3 

4 

5 

7 

8 

9 

11 

12 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

3 

4 

5 

6 

8 

9 

10 

12 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

3 

4 

5 

6 

8 

9 

10 

11 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

2 

4 

5 

6 

7 

9 

10 

11 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

2 

4 

5 

6 

7 

8 

10 

11 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

2 

3 

5 

6 

7 

8 

9 

10 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

2 

3 

5 

6 

7 

8 

9 

10 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

2 

3 

4 

5 

7 

8 

9 

10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

2 

3 

4 

5 

6 

8 

9 

10 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

2 

3 

4 

5 

6 

7 

8 

9 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

2 

3 

4 

5 

6 

7 

8 

9 

43 

6335 

6345 

6355 

6365 

3375 

6385 

6395 

6405  6415 

6425 

2 

3 

4 

5 

6 

7 

8 

9 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

2 

3 

4 

5 

6 

7 

8 

9 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

2 

3 

4 

5 

6 

7 

8 

9 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

2 

3 

4 

5 

6 

7 

7 

8 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

2 

3 

4 

5 

5 

6 

7 

8 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

2 

3 

4 

4 

5 

6 

7 

8 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

2 

3 

4 

4 

5 

6 

7 

8 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

2 

3 

3 

4 

5 

6 

7 

8 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

2 

3 

3 

4 

5 

6 

7 

8 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

2 

o 

3 

4 

5 

6 

7 

7 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

2 

2 

3 

4 

5 

6 

6 

7 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

2 

2 

3 

4 

5 

6 

6 

7 

LOGARITHMS  OF  NUMBERS. 


199 


^2 

li 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Proportional  Parts. 

1 

2 

8 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7469 

7466 

7474 

2 

2 

3 

5 

5 

6 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

2 

2 

3 

5 

5 

6 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

2 

2 

3 

5 

5 

6 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

2 

3 

5 

6 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

2 

3 

5 

6 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

2 

3 

5 

6 

6 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

2 

3 

5 

6 

6 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

2 

3 

3 

5 

6 

6 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

2 

3 

3 

5 

5 

6 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

2 

3 

3 

5 

5 

6 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

2 

3 

3 

5 

5 

6 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

2 

3 

3 

5 

5 

6 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

2 

3 

3 

5 

5 

6 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

2 

3 

3 

5 

6 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

2 

2 

3 

5 

6 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

2 

2 

3 

5 

6 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

2 

2 

3 

5 

5 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

2 

2 

3 

5 

5 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

2 

2 

3 

5 

5 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

2 

2 

3 

5 

5 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

2 

2 

3 

3 

5 

5 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

1 

2 

2 

3 

3 

5 

5 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

2 

2 

3 

3 

5 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

2 

2 

3 

3 

5 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

2 

2 

3 

3 

5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

2 

2 

3 

3 

5 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

2 

2 

3 

3 

5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

2 

2 

3 

3 

6 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

2 

2 

3 

3 

5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

2 

2 

3 

3 

5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

2 

2 

3 

3 

5 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

2 

2 

3 

3 

5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

0 

2 

2 

3 

3 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

0 

1 

2 

2 

3 

3 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

0 

^ 

2 

2 

3 

3 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

0 

2 

2 

3 

3 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

0 

2 

2 

3 

3 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

0 

2 

2 

3 

3 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

0 

2 

2 

3 

3 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

0 

2 

2 

3 

3 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

0 

2 

2 

3 

3 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

0 

2 

2 

3 

3 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

0 

2 

2 

3 

3 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

0 

2 

2 

3 

3 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

0 

2 

2 

3 

3 

JL 

200 


ANTILOGARITHMS. 


ii 

Proportional  Parts. 

&"_= 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

"^•S 

-^•c 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.00 

1000 

1002 

1005 

1007 

1009 

1012 

1014 

1016 

1019 

1021 

0 

0 

1 

1 

2 

2 

2 

.01 

1023 

1026 

1028 

1030 

1033 

1035 

1038 

1040 

1042 

1045 

0 

0 

1 

1 

2 

2 

2 

.02 

1047 

1050 

1052 

1054 

1057 

1059 

1062 

1064 

1067 

1069 

0 

0 

1 

1 

2 

2 

2 

.03 

1072 

1074 

1076 

1079 

1081 

1084 

1086 

1089 

1091 

1094 

0 

0 

1 

1 

2 

2 

2 

.04 

1096 

1099 

1102 

1104 

1107 

1109 

1112 

1114 

1117 

1119 

0 

1 

1 

2 

2 

2 

2 

.05 

1122 

1125 

1127 

1130 

1132 

1135 

1138 

1140 

1143 

1146 

0 

1 

2 

2 

2 

2 

.06 

1148 

1151 

1153 

1156 

1159 

1161 

1164 

1167 

1169 

1172 

0 

1 

2 

2 

2 

2 

,07 

1175 

1178 

1180 

1183 

1186 

1189 

1191 

1194 

1197 

1199 

0 

1 

2 

2 

2 

2 

.08 

1202 

1205 

1208 

1211 

1213 

1216 

1219 

1222 

1225 

1227 

0 

1 

2 

2 

2 

3 

.09 

1230 

1233 

1236 

1239 

1242 

1245 

1247 

1250 

1253 

1256 

0 

1 

2 

2 

2 

3 

.10 

1259 

1262 

1265 

1268 

1271 

1274 

1276 

1279 

1282 

1285 

0 

1 

2 

2 

2 

3 

.11 

1288 

1291 

1294 

1297 

1300 

1303 

1306 

1309 

1312 

1315 

0 

2 

2 

2 

2 

3 

.12 

1318 

1321 

1324 

1327 

1330 

1334 

1337 

1340 

1343 

1346 

0 

2 

2 

•> 

2 

3 

.13 

1349 

1352 

1355 

1358 

1361 

1365 

1368 

1371 

1374 

1377 

0 

2 

2 

2 

3 

3 

.14 

1380 

1384 

1387 

1390 

1393 

1396 

1400 

1403 

1406 

1409 

0 

2 

2 

2 

3 

3 

.15 

1413 

1416 

1419 

1422 

1426 

1429 

1432 

1435 

1439 

1442 

0 

2 

2 

2 

3 

3 

.16 

1445 

1449 

1452 

1455 

1459 

1462 

1466 

1469 

1472 

1476 

0 

2 

2 

2 

3 

3 

.17 

1479 

1483 

1486 

1489 

1493 

1496 

1500 

1503 

1507 

1510 

0 

2 

2 

2 

3 

3 

.18 

1514 

1517 

1521 

1524 

1528 

1531 

1535 

1538 

1542 

1545 

0 

2 

2 

2 

3 

3 

.19 

1549 

1552 

1556 

1560 

1563 

1567 

1570 

1574 

1578 

1581 

0 

2 

2 

3 

3 

3 

.20 

1585 

1589 

1592 

1596 

1600 

1603 

1607 

1611 

1614 

1618 

0 

1 

1 

2 

2 

3 

3 

3 

.21 

1622 

1626 

1629 

1633 

1637 

1641 

1644 

1648 

1652 

1656 

0 

2 

2 

2 

3 

3 

3 

.22 

1660 

1663 

1667 

1671 

1675 

1679 

1683 

1687 

1690 

1694 

0 

2 

2 

2 

3 

3 

3 

.23 

1698 

1702 

1706 

1710 

1714 

1718 

1722 

1726 

1730 

1734 

0 

2 

2 

2 

3 

3 

4 

.24 

1738 

1742 

1746 

1750 

1754 

1758 

1762 

1766 

1770 

1774 

0 

2 

2 

2 

3 

3 

4 

.25 

1778 

1782 

1786 

1791 

1795 

1799 

1803 

1807 

1811 

1816 

0 

2 

2 

2 

3 

3 

4 

.26 

1820 

1824 

1828 

1832 

1837 

1841 

1845 

1849 

1854 

1858 

0 

2 

2 

3 

3 

3 

4 

.27 

1862 

1866 

1871 

1875 

1879 

1884 

1888 

1892 

1897 

1901 

0 

2 

2 

3 

3 

3 

4 

.28 

1905 

1910 

1914 

1919 

1923 

1928 

1932 

1936 

1941 

1945 

0 

2 

2 

3 

3 

4 

4 

.29 

1950 

1954 

1959 

1963 

1968 

1972 

1977 

1982 

1986 

1991 

0 

2 

2 

3 

3 

4 

4 

.30 

1995 

2000 

2004 

2009 

2014 

2018 

2023 

2028 

2032 

2037 

0 

2 

2 

3 

3 

4 

4 

.31 

2042 

2046 

2051 

2056 

2061 

2065 

2070 

2075 

2080 

2084 

0 

2 

2 

3 

3 

4 

4 

.32 

2089 

2094 

2099 

2104 

2109 

2113 

2118 

2123 

2128 

2133 

0 

2 

2 

3 

3 

4 

4 

.33 

2138 

2143 

2148 

2153 

2158 

2163 

2168 

2173 

2178 

2183 

0 

1 

2 

2 

3 

3 

4 

4 

.34 

2188 

2193 

2198 

2203 

2208 

2213 

2218 

2223 

2228 

2234 

1 

2 

2 

3 

3 

4 

4 

5 

.35 

2239 

2244 

2249 

2254 

2259 

2265 

2270 

2275 

2280 

2286 

2 

2 

3 

3 

4 

4 

5 

.36 

2291 

2296 

2301 

2307 

2312 

2317 

2323 

2328 

2333 

2339 

2 

2 

3 

3 

4 

4 

5 

.37 

2344 

2350 

2355 

2360 

2366 

2371 

2377 

2382 

2388 

2393 

2 

2 

3 

3 

4 

4 

5 

.38 

2399 

2404 

2410 

2415 

2421 

2427 

2432 

2438 

2443 

2449 

2 

2 

3 

3 

4 

4 

5 

.39 

2455 

2460 

2466 

2472 

2477 

2483 

2489 

2495 

2500 

2506 

2 

2 

3 

3 

4 

5 

5 

.40 

2512 

2518 

2523 

2529 

2535 

2541 

2547 

2553 

2559 

2564 

2 

2 

3 

4 

4 

5 

5 

.41 

2570 

2576 

2582 

2588 

2594 

2600 

2606 

2612 

2618 

2624 

2 

2 

3 

4 

4 

5 

5 

.42 

2630 

2636 

2642 

2649 

2655 

2661 

2667 

2673 

2679 

2685 

2 

2 

3 

4 

4 

5 

6 

.43 

2692 

2698 

2704 

2710 

2716 

2723 

2729 

2735 

2742 

2748 

2 

3 

3 

4 

4 

5 

6 

.44 

2754 

2761 

2767 

2773 

2780 

2786 

2793 

2799 

2805 

2812 

2 

3 

3 

4 

4 

5 

6 

.45 

2818 

2825 

2831 

2838 

2844 

2851 

2858 

2864 

2871 

2877 

2 

3 

3 

4 

5 

5 

6 

.46 

2884 

2891 

2897 

2904 

2911 

2917 

2924 

2931 

2938 

2944 

2 

3 

3 

4 

5 

5 

6 

.47 

2951 

2958 

2965 

2972 

2979 

2985 

2992 

2999 

3006 

3013 

2 

3 

3 

4 

5 

5 

6 

.48 

3020 

13027 

3034 

3041 

3048 

3055 

3062 

3069 

3076 

3083 

2 

3 

4 

4 

5 

6 

6 

.49 

3090 

13097 

I31O5 

3112 

3119 

3126 

3133  3141 

3148 

3155 

2 

3 

4 

4 

5 

6 

6 

ANTILOGARITHMS. 


201 


li 

,. 

Proportional  Parts. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

^1 

— 

— 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.50 

3162 

3170 

317.7 

3184 

3192 

3199 

3206 

3214 

3221 

3228 

1 

2 

3 

4 

4 

5 

6 

.51 

3236! 

3243' 

3251 

3258 

3266 

3273 

3281 

3289 

32961 

3304 

2 

2 

3 

4 

5 

5 

6 

.52 

3311! 

3319 

3327 

33341 

3342 

3350 

3357 

3365! 

3373] 

3381 

1 

2 

2 

3 

4 

5 

5 

6 

.53 

3388 

3396 

3404 

3412 

3420 

3428; 

3436 

3443 

3451 

3459 

2 

2 

3 

4 

6 

6 

6 

.54 

3467 

3475 

3483 

3491 

3499 

3508 

3516 

3524 

3532 

3540 

2 

2 

3 

4 

5 

6 

6 

.55 

3548 

3556 

3565 

3573 

3581 

3589 

3597 

3606 

3614 

3622 

2 

2 

3 

4 

5 

6 

7 

.56 

3631 

3639 

3648 

3656 

3664 

3673 

3681 

3690 

3698 

3707 

2 

3 

3 

4 

5 

6 

8 

.57 

3715 

3724 

3733 

3741 

3750 

3758 

3767 

3776 

3784 

3793 

2 

3 

3 

4 

5 

6 

8 

.58 

3802 

3811 

3819 

3828 

3837 

3846 

3855 

3864 

3873 

3882 

2 

3 

4 

4 

5 

6 

8 

.59 

3890 

3899 

3908 

3917 

3926 

3936 

3945 

3954 

3963 

3972 

2 

3 

4 

5 

5 

6 

8 

.60 

3981 

3990 

3999 

4009 

4018 

4027 

4036 

4046 

4055 

4064 

2 

3 

4 

5 

6 

6 

7 

8 

.61 

4074 

4083 

4093 

4102 

4111 

4121 

4130 

4140 

4150 

4159 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.62 

4169 

4178 

4188 

4198 

4207 

4217 

4227 

4236 

4246 

4256 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.63 

4266 

4276 

4285 

4295 

4305 

4315 

4325 

4335 

4345 

4355 

1 

2 

3 

4 

5 

6 

<" 

8 

9 

.64 

4365 

4375 

4385 

4395 

4406 

4416 

4426 

4436 

4446 

4457 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.65 

4467 

4477 

4487 

4498 

4508 

4519 

4529 

4539 

4550 

4560 

2 

3 

4 

5 

6 

7 

8 

9 

M 

4571 

4581 

4592 

4603 

4613 

4624 

4634 

4645 

4656 

4667 

2 

3 

4 

5 

6 

7 

9 

10 

.67 

4677 

4688 

4699 

4710 

4721 

4732 

4742 

4753 

4764 

4775 

2 

3 

4 

5 

7 

8 

9 

10 

.68 

4786 

4797 

4808 

4819 

4831 

4842 

4853 

4864 

4875 

4887 

2 

3 

4 

6 

7 

8 

9 

10 

.69 

4898 

4909 

4920 

4932 

4943 

4955 

4966 

4977 

4989 

5000 

2 

3 

5 

6 

7 

8 

9 

10 

.70 

5012 

5023 

5035 

5047 

5058 

5070 

5082 

5093 

5105 

5117 

2 

4 

5 

6 

7 

8 

9 

11 

.71 

5129 

5140 

5152 

5164 

5176 

5188 

5200 

5212 

5224 

5236 

2 

4 

5 

6 

7 

8 

10 

11 

.72 

5248 

5260 

5272 

5284 

5297 

5309 

5321 

5333 

5346 

5358 

2 

4 

5 

6 

7 

9 

10 

11 

.73 

5370 

5383 

5395 

5408 

5420 

5433 

5445 

5458 

5470 

5483 

3 

4 

5 

6 

8 

9 

10 

11 

.74 

5495 

5508 

5521 

5534 

5546 

5559 

5572 

5585 

5598 

5610 

3 

4 

6 

6 

8 

9 

10 

12 

.75 

5623 

5636 

5649 

5662 

5675 

5689 

5702 

5715 

5728 

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DATE  DUE  SLIP 

UNIVERSITY  OF  CALIFORNIA  MEDICAL  SCHOOL  LIBRARY 

THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


MAR  27  1939     ' 
.CICT2  71952 


gg.'6-«^Z 

A 

^^k 

^^H 

^^^^1 

^^^^^1 

■ 

^^^^^^ 

^^^^^r 

